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Given a string of characters, we can permute the individual characters to make new strings. If we can impose an ordering on the characters (say alphabetic sequence), then the strings themselves can be ordered and any given permutation can be given a unique number designating its position in that ordering. For example the string `acab' gives rise to the following 12 distinct permutations:

Thus the string `acab' can be characterised in this sequence as 5.

Write a program that will read in a string and determine its position in the ordered sequence of permutations of its constituent characters. Note that numbers of permutations can get very large; however we guarantee that no string will be given whose position is more than .

Input and Output

Input will consist of a series of lines, each line containing one string. Each string will consist of up to 30 lower case letters, not necessarily distinct. The file will be terminated by a line consisting of a single #.

Output will consist of a series of lines, one for each line of the input. Each line will consist of the position of the string in its sequence, right justified in a field of width 10.

Sample input

bacaa
abc
cba
#

Sample output

        15
         1
         6

bacaa 當要計算這個的時候, 'b' 前面還有幾個字母, 嘗試做為頭
則 可以 'a' 開頭 計算排列 aacb 的個數
'ba' 已經是底線了,
'bac' 可以 'a' 開頭, 計算 aac 的個數 ... 類推

#include <stdio.h>
#include <string.h>
int gcd(int x, int y) {
    int t;
    while(x%y) {
        t = x, x = y, y = t%y;
    }
    return y;
}
int main() {
    char s[50];
    while(scanf("%s", s) == 1) {
        if(s[0] == '#')
            break;
        int cnt[26] = {}, len = strlen(s);
        int i, j, k, l, a;
        for(i = 0; s[i]; i++) {
            cnt[s[i]-'a']++;
        }
        long long ans = 0;
        for(i = 0; s[i]; i++) {
            for(j = 0; j < s[i]-'a'; j++) {
                if(cnt[j] == 0) continue;
                cnt[j]--;
                int set[50] = {};
                for(k = 2; k < len-i; k++)
                    set[k] = k;
                for(k = 0; k < 26; k++) {
                    for(l = 2; l <= cnt[k]; l++) {
                        int tmp = l;
                        for(a = 2; a < len-i; a++) {
                            int g = gcd(tmp, set[a]);
                            tmp /= g;
                            set[a] /= g;
                        }
                    }
                }
                long long tmp = 1;
                for(k = 2; k < len-i; k++)
                    tmp *= set[k];
                ans += tmp;
                cnt[j]++;
            }
            cnt[s[i]-'a']--;
        }
        printf("%10lld\n", ans+1);
    }
    return 0;
}