e.g Let's consider aabb. shifting the rightmost 'a' to the end of the string will take two operations but it will double the number of b's (abbbba) then shifting the leftmost 'a' past the b's will cost the new number of b's but also increase the new number of b's by two (bbbbbbbbaa). from this we can see that it's a geometric progression with common ratio of 2, geometric sum formular is $$$\frac{a(r^2-1)}{(r-1)}$$$ where a is the first term and r is the common ratio. so we can just apply this : $$$\frac{b_{c}(a_{c}^2-1)}{(2-1)}$$$ == $$$b_{c}(a_{c}^2-1)$$$ to find answer of a substring of form aaaa.....aabbbb....bb. where $$$a_{c}$$$ and $$$b_{c}$$$ are the count of a's and b's respectively.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000007
long long binpow(long long a, long long b){
	long long res = 1;
	while(b>0){
		if(b&1) res = (res * a)%MOD;
		a = (a*a)%MOD;
		b >>= 1;
	}
	return res%MOD;
}
int main(){
	
	string s; cin>>s;
	ll ans = 0; ll a = 0; ll b = 0;
	for(int i = 0; i < s.size(); i++){
		if(i-1 >= 0) if(s[i] == 'a' && s[i-1] == 'b') {
			ans += (b*(binpow(2,a)-1));
			b = 0;
		}
		if(s[i] == 'a') a++;
		else b++;
		ans %= MOD;
	}
	ans += (b*(binpow(2,a)-1));
	ans %= MOD;
	cout<<ans<<"\n";
	
}