Codeforces Round #411(Div. 2)D. Minimum number of steps【思维递推+快速幂】

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 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".

题目大意：

给你一个只包含a，b两种字母的字符串，如果有一个连续子串为ab.那么就可以将其变成bba.让你计算这个给出的字符串在操作了多少次之后不能再进行操作了。

思路：

最终字符串一定会变成bbbbbbb........aaaa..............

所以前边的a是要累加起来的。

我们观察到ab答案是1

aab答案是3.

aaab答案是7.

那么一个b前边有n个a，那么这个b会贡献2^n-1次操作。

abab答案是4 =1 + 3

那么对应累加前边的a.一旦遇到b就加上2^n-1即可。

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
#define mod 1000000007
char a[1000800];
ll poww(ll a,ll b)
{
    ll tmp=a%mod;
    ll ans=1;
    ll n=b;
    while(n)
    {
        if(n%2==1)
        {
            ans=(ans*tmp)%mod;
            n-=1;
        }
        else
        {
            tmp=(tmp*tmp)%mod;
            n/=2;
        }
    }
    return ans;
}
int main()
{
    while(~scanf("%s",a))
    {
        ll cnt=0;
        ll output=0;
        ll n=strlen(a);
        for(ll i=0;i<n;i++)
        {
            if(a[i]=='b')
            {
                output=(output+(poww(2,cnt)-1)+mod)%mod;
            }
            else cnt++;
        }
        printf("%I64d\n",(output+mod)%mod);
    }
}