The graveyard is organized as a grid of W x H cells, with the entrance in the cell at position (0, 0) and the exit at (W - 1, H - 1). Despite the darkness, Scared John can always recognize the exit, and he will leave the moment he reaches it, determined never to set foot anywhere in the graveyard again. On his way to the exit, he can walk from one cell to an adjacent one, and he can only head to the North, East, South or West. In each cell there can be either one gravestone, one ``haunted hole'', or grass:

• If the cell contains a gravestone, you cannot walk over it, because gravestones are too high to climb.

• If the cell contains a ``haunted hole'' and you reach it, you will appear somewhere in the graveyard at a possibly different moment in time. The time difference depends on the particular ``haunted hole'' you fell into, and can be positive, negative or zero.

• Otherwise, the cell has only grass, and you can walk freely over it.

He is terrified, so he wants to cross the graveyard as quickly as possible. And that is the reason why he has phoned you, a renowned programmer. He wants you to write a program that, given the description of the graveyard, computes the minimum time needed to go from the entrance to the exit. Scared John accepts using ``haunted holes'' if they permit him to cross the graveyard quicker, but he is frightened to death of the possibility of getting lost and being able to travel back in time indefinitely using the holes, so your program must report these situations.

Figure 3 illustrates a possible graveyard (the second test case from the sample input). In this case there are two gravestones in cells (2, 1) and (3, 1), and a ``haunted hole'' from cell (3, 0) to cell (2, 2) with a difference in time of 0 seconds. The minimum time to cross the graveyard is 4 seconds, corresponding to the path:

(0, 0)(1, 0)(2, 0)(3, 0)(2, 2)(3, 2)

Input 

The input consists of several test cases. Each test case begins with a line containing two integers W and H ( 1W, H 30). These integers represent the width W and height H of the graveyard. The next line contains an integer G (G 0), the number of gravestones in the graveyard, and is followed by G lines containing the positions of the gravestones. Each position is given by two integers X and Y ( 0 X < W and 0 Y < H).

The next line contains an integer E (E 0), the number of ``haunted holes'', and is followed by E lines. Each of these contains five integers X1, Y1, X2, Y2, T. (X1, Y1) is the position of the ``haunted hole'' ( 0 X1 < W and 0 Y1 < H). (X2, Y2) is the destination of the ``haunted hole'' ( 0 X2 < W and 0 Y2 < H). Note that the origin and the destination of a ``haunted hole'' can be identical. T ( -10 000 T 10 000) is the difference in seconds between the moment somebody enters the ``haunted hole'' and the moment he appears in the destination position; a positive number indicates that he reaches the destination after entering the hole. You can safely assume that there are no two ``haunted holes'' with the same origin, and the destination cell of a ``haunted hole'' does not contain a gravestone. Furthermore, there are neither gravestones nor ``haunted holes'' at positions (0,0) and (W - 1, H - 1).

The input will finish with a line containing `0 0', which should not be processed.

Output 

Sample Input 

3 3
2
2 1
1 2
0
4 3
2
2 1
3 1
1
3 0 2 2 0
4 2
0
1
2 0 1 0 -3
0 0

Sample Output 

Impossible
4
Never

這膽小鬼要經過墓園，其中墓碑是無法經過得地方。

其中有幾處地方有墓穴，會掉入異次元跑到另一個地方，中間會存在不確定的時間流逝。

由於這個膽小鬼想要盡可能走出墓園，因此它會盡可能走墓穴，有可能會無限輪迴於這個墓穴傳遞。

一旦經過墓穴，一定會被吸進去！

膽小鬼只能上下左右走，每走一步消耗 1 秒。

問 1. 困在墓穴(不斷地回朔至過去) 2. 根本離不開 3. 最小消耗

題目解法：

使用 SPFA 找負環。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <deque>
using namespace std;
int g[35][35];
int hole[35][35][3];
int W, H, G, E;
int spfa(int &negcycle) {
	deque<int> X, Y;
	int x, y, tx, ty;
	int i, j, k;
	int dx[] = {0,0,1,-1};
	int dy[] = {1,-1,0,0};
	int inq[35][35] = {}, inqc[35][35] = {};
	int dist[35][35];
	int node = 0;
	for(i = 0; i < W; i++)
		for(j = 0; j < H; j++)
			node += g[i][j] == 0;
	memset(dist, 63, sizeof(dist));
	X.push_front(0), Y.push_front(0);
	dist[0][0] = 0, inqc[0][0]++;
	while(!X.empty()) {
		x = X.front(), X.pop_front();
		y = Y.front(), Y.pop_front();
		inq[x][y] = 0;
		if(x == W-1 && y == H-1)
			continue;
		if(hole[x][y][0] != -1) {//haunted hole
			tx = hole[x][y][0], ty = hole[x][y][1];
			if(dist[tx][ty] > dist[x][y] + hole[x][y][2]) {
				dist[tx][ty] = dist[x][y] + hole[x][y][2];
				if(inq[tx][ty] == 0)  {
					if(!X.empty() && dist[X.front()][Y.front()] > dist[tx][ty])
						X.push_front(tx), Y.push_front(ty);
					else
						X.push_back(tx), Y.push_back(ty);
					inq[tx][ty] = 1;
					if(++inqc[tx][ty] > node) {
						negcycle = 1;
						return 0; // negative cycle
					}
				}
			}
			continue;//important
		}
		for(i = 0; i < 4; i++) {
			tx = x+dx[i], ty = y+dy[i];
			if(tx < 0 || ty < 0 || tx >= W || ty >= H)
				continue;
			if(g[tx][ty])	continue;//gravestone
			if(dist[tx][ty] > dist[x][y]+1) {
				dist[tx][ty] = dist[x][y]+1;
				if(inq[tx][ty] == 0)  {
					if(!X.empty() && dist[X.front()][Y.front()] > dist[tx][ty])
						X.push_front(tx), Y.push_front(ty);
					else
						X.push_back(tx), Y.push_back(ty);
					inq[tx][ty] = 1;
					if(++inqc[tx][ty] > node) {
						negcycle = 1;
						return 0; // negative cycle
					}
				}
			}
		}
	}
	return dist[W-1][H-1];
}
int main() {
	int i, j, k, x, y;
	int sx, sy, ex, ey, t;
	while(scanf("%d %d", &W, &H) == 2 && W+H) {
		memset(g, 0, sizeof(g));
		memset(hole, -1, sizeof(hole));
		scanf("%d", &G);
		for(i = 0; i < G; i++) {
			scanf("%d %d", &x, &y);
			g[x][y] = 1;
		}
		scanf("%d", &E);
		for(i = 0; i < E; i++) {
			scanf("%d %d %d %d %d", &sx, &sy, &ex, &ey, &t);
			hole[sx][sy][0] = ex;
			hole[sx][sy][1] = ey;
			hole[sx][sy][2] = t;
		}
		int negcycle = 0;
		int ret = spfa(negcycle);
		if(negcycle)
			puts("Never");
		else if(ret == 0x3f3f3f3f)
			puts("Impossible");
		else
			printf("%d\n", ret);
	}
	return 0;
}