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# include <bits/stdc++.h>
using namespace std ;
string str ;
int ans ;
//求组合数的优化后办法
long long C ( int n , int m ) {
long long sum = 1 ;
for ( int i = 1 ; i < = m ; i + + )
sum = sum * ( n - m + i ) / i ;
return sum ;
}
/**
* 测试用例：
* 答案：
* @return
*/
int main ( ) {
//输入字符串
cin > > str ;
//字符串长度
int len = str . size ( ) ;
//判断是否合法
for ( int i = 0 ; i < len ; i + + )
if ( str [ i ] < = str [ i - 1 ] ) { //非升序就是不正确的编码
cout < < 0 ;
exit ( 0 ) ;
}
//1、计算出位数比它小的单词数
//比如是cgx,那么就计算出
// 1位的有C(26,1)个
// 2位的有C(26,2)个
for ( int i = 1 ; i < len ; i + + ) ans + = C ( 26 , i ) ;
//2、到这就是位数和它自己一样的了
// 枚举每一位,从低到高，一个个来
for ( int pos = 0 ; pos < len ; pos + + ) {
//每一位都要找到比当前输入字符串小的字符串个数，叠加在一起就是结果
char start = ' a ' ; //初值为`a`,首位的话，从`a`开始，其它的从上一位的后面一个开始
if ( pos > 0 ) start = str [ pos - 1 ] + 1 ; //str的数组下标是从0开始的
for ( char j = start ; j < str [ pos ] ; j + + ) //注意范围,找比它小的
ans + = C ( ' z ' - j , len - pos - 1 ) ; //'z' - j 剩下可用的字符数量, len-pos-1需要选择的字符数量
}
//别忘了最后把自己加上
cout < < + + ans ;
return 0 ;
}
