CodeFroces 804B Minimum number of steps

B. Minimum number of steps

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 109 + 7.

The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.

Input

The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 106.

Output

Print the minimum number of steps modulo 109 + 7.

Examples

input

ab

output

1

input

aab

output

3

Note

The first example: "ab"  →  "bba".

The second example: "aab"  →  "abba"  →  "bbaba"  →  "bbbbaa".

题意：给出一个字符串，其中ab的地方用bba代替，直到没有ab为止，问要操作多少次，答案对1e9+7取模。

我是按照ab前面有多少个a开始计数的。

手写几个发现，ab前面0个a的时候输出1,1个a的时候输出3,2个a的时候输出7,3个a的时候输出15，很容易发现这其实是2^(n+1)-1。

还发现，前面有n个a到了后面就有n+1个a。比如abab变成bbaab，后面的ab前面有1个前面转化过来的a。

最后还要发现，abbbb，那么就会生成4个ab，而且他们是按照最前面那个ab的前面有多少个a来计数的。

然后，打表，输出，即可。

代码如下：

#include<bits/stdc++.h>using namespace std;const int mod = 1e9 + 7;const int maxn = 1e6 + 5;char str[maxn];int cnt[maxn];void init(){cnt[0] = 1;for(int i = 1; i < maxn; i++){cnt[i] = ( ( (cnt[i - 1] % mod) * 2) % mod + 1) % mod;}}int main(){int c = 0;int ans = 0;bool flag = 0;init();scanf("%s", &str);for(int i = 0; str[i]; i++){if(flag){if(str[i] == 'b'){ans = ((ans % mod) + (cnt[c] % mod)) % mod;continue;}else{flag = 0;c++;}}if(str[i] == 'a' && str[i + 1] == 'b'){ans = ((ans % mod) + (cnt[c] % mod)) % mod;i = i + 1;flag = 1;}else if(str[i] == 'a')c++;}cout << ans << endl;return 0;}