CodeForces - 805D 思維

題目：當遇到ab時候變成bba，步數加一，最後輸出步數

思路：所有的a到最後都是從左到右，而且a的數量不變

           之後ab後邊有b就一直加一，變換後b的數量也會加一

            做法呢，就是從右邊開始遍歷，遇到a後就叫遇到的b乘以2

#include<bits/stdc++.h>
#define m 1000000007
char ch[1000002];
int main()
{
	
	while(scanf("%s",ch)!=EOF)
	{
		long long cnt=0,ans=0;
		int l=strlen(ch);
		for(int i=l-1;i>=0;i--)
		{
			if(ch[i]=='b')
			{
				cnt++;
			}
			else if(ch[i]=='a') 
			{
				ans=(ans+cnt)%m;
				cnt=(cnt*2)%m;
			}
		}
		printf("%lld\n",ans);
	}
}

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".