#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<(to);x++)
#define FORR(x,arr) for(auto& x:arr)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

ll dp[202020][27];
const ll mo=1000000007;
string S;
ll p2[202020];

void solve() {
	int i,j,k,l,r,x,y; string s;
	
	dp[0][26]=1;
	
	p2[0]=1;
	FOR(i,201010) p2[i+1]=p2[i]*2%mo;
	
	ll ret=0;
	
	cin>>S;
	FOR(i,S.size()) {
		FOR(j,27) {
			// keep
			(dp[i+1][j]+=dp[i][j])%=mo;
			// add
			(dp[i+1][S[i]-'a']+=dp[i][j])%=mo;
			if(j!=S[i]-'a' && j!=26) {
				(ret+=dp[i][j]*p2[S.size()-1-i])%=mo;
			}
		}
	}
	FOR(i,26) ret+=dp[S.size()][i];
	cout<<ret%mo<<endl;
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false), cin.tie(0);
	FOR(i,argc-1) s+=argv[i+1],s+='\n'; FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	cout.tie(0); solve(); return 0;
}