CodeForces - 804BMinimum number of steps（贪心）

Description

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，ab不能出现在这个字符串中，ab可以变成bba，求出最少的操作次数，使字符串合法。

ab变成bba相当于ab倒置然后在最前面加上一个b，把ab变成bba会影响前面的字符所以最佳的方法是从后向前遍历。

遇到b就+1.，变换后相当于又加了一个b，所以当遇到a时把前面变换的次数乘2.

AC代码：

#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define INF 0x3f3f3f3f typedef long long ll; const int m=1e9+7; char a[1000010]; int main() { gets(a); int len=strlen(a); int i,j,k,l; int sum=0,ans=0; for(i=len-1;i>=0;i--) { if(a[i]=='b') sum=(sum+1)%m; else { ans=(ans+sum)%m; sum=(sum*2)%m; //cout<<ans<<endl; } } printf("%d\n",ans); return 0; }