Minimum number of steps

D. 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 modulo10⁹ + 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 from1 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".

本题题意：对于只有a和b的一个字符串，问只要所给的串中有子串中有ab,就将ab换为bba直到得的字符串不含ab子串为止，输出对于变换的最小次数；

理解题意及找规律知，本题就是将字符串中的a全都移到最右边就是所得结果，每次移动a个数不变，b个数加一；操作步骤就是从右到左把a移到右边，每次移动通过找规律知：

每次移动当前a的变换次数，就是将这个a移到右端时，它右边的b的个数。注意：每次已过之后，会发现，b的个数翻倍了。

#include<iostream>#include<stdio.h>using namespace std;int main(){    string str;    while(cin >> str)    {    long long n = str.length();    long long sum = 0, i, k = 0, ans = 0;    for(i = n - 1; i >= 0; i--)    {        if(str[i] == 'a')        {            ans = ((ans * 2 % 1000000007)+ k % 1000000007)% 1000000007;  //注意这个地方也要求余；不然会超出范围，这是两次错误后找到的bug....            sum = (sum % (1000000007) + ans % (1000000007)) % (1000000007) ;            k = 0;            continue;        }        k++;    }    cout << sum % (1000000007)<< endl;    }    return 0;}