Given a non-empty string composed of digits only, we may group these digits into sub-groups (but maintaining their original order) if, for every sub-group but the last one, the sum of the digits in a sub-group is less than or equal to the sum of the digits in the sub-group immediately on its right. Needless to say, each digit will be in exactly one sub-group.

For example, the string 635 can only be grouped in one sub-group [635] or in two sub-groups as follows: [6-35] (since 6 < 8.) Another example is the string 1117 which can be grouped in one sub-group [1117] or as in the following: [1-117], [1-1-17], [1-11-7], [1-1-1-7], [11-17] and [111-7] but not any more, hence the total number of possibilities is 7.

Write a program that computes the total number of possibilities of such groupings for a given string of digits.

Input

Your program will be tested on a number of test cases. Each test case is specified on a separate line. Each line contains a single string no longer than 25, and is made of decimal digits only.

The end of the test cases is identified by a line made of the word "bye" (without the quotes.) Such line is not part of the test cases.

Output

For each test case, write the result using the following format:

k. n

where k is the test case number (starting at 1,) and n is the result of this test case.

Example

Input:

635

1117

9876

bye

Output:

1. 2

2. 7

3. 2
      f[i][j]表示进行到第i位，且最后一个集合包含j个数字的方案个数，显然f[i][j]=SUM{f[i-j][k]  | 只要最后i的最后j个数的和>=i-j的最后k个数的和就能转移}

 #include<bits/stdc++.h>

 using namespace std;

 #define inf 0x3f3f3f3f

 int main()

 {

     char s[];

     int x[]={};

     int f[][];

     int i,j,k,cas=;

     while(scanf("%s",s+)){

         if(!strcmp(s+,"bye")) break;

         int n=strlen(s+);

         for(i=;i<=n;++i){

             x[i]=x[i-]+s[i]-'';

         }

         memset(f,,sizeof(f));

         f[][]=f[][]=;

         for(i=;i<=n;++i){

             for(j=;j<=i;++j){

                 if(i==j) f[i][j]=;

                 else f[i][j]=;

                 for(k=;k<=i-j;++k){

                     if(x[i]-x[i-j]>=x[i-j]-x[i-j-k])

                      f[i][j]+=f[i-j][k];

                 }

             }

         }

         int s=;

         for(i=;i<=n;++i) s+=f[n][i];

         cout<<++cas<<". "<<s<<endl;

     }

     return ;

 }