#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define P pair<ll,ll>
#define FOR(I,A,B) for(ll I = ll(A); I < ll(B); ++I)
#define FORR(I,A,B) for(ll I = ll((B)-1); I >= ll(A); --I)
#define TO(x,t,f) ((x)?(t):(f))
#define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin()) //xi>=v  x is sorted
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin()) //xi>v  x is sorted
#define NUM(x,v) (POSU(x,v)-POSL(x,v))  //x is sorted
#define REV(x) (reverse(x.begin(),x.end())) //reverse
ll gcd_(ll a,ll b){if(a%b==0)return b;return gcd_(b,a%b);}
ll lcm_(ll a,ll b){ll c=gcd_(a,b);return ((a/c)*(b/c)*c);}
#define NEXTP(x) next_permutation(x.begin(),x.end())
const ll INF=ll(1e16)+ll(7);
const ll MOD=1000000007LL;
#define out(a) cout<<fixed<<setprecision((a))
//tie(a,b,c) = make_tuple(10,9,87);
#define pop_(a) __builtin_popcount((a))
ll keta(ll a){ll r=0;while(a){a/=10;r++;}return r;}
ll ketawa(ll a){ll r=0;while(a){r+=a%10;a/=10;}return r;}



int main(){

  string s;
  cin >> s;
  vector<ll> X(26,0),Y(26,0);
  ll N = s.size();
  ll ans = 0,ans2=0;
  FOR(i,0,N){
    ll c = s[i] - 'a';
    FOR(j,0,26){
      if(c==j){
        ( ans += X[j] ) %= MOD;
      }else{
        ( ans += X[j] + Y[j] ) %= MOD;
      }
    }
    ( ans += 1 ) %= MOD;
    (X[c]*=2)%=MOD;
    (X[c]+=1)%=MOD;
    FOR(j,0,26)if(j!=c)(X[c]+=Y[j]+X[j])%=MOD;
    (Y[c]*=2)%=MOD;
    (Y[c]+=1)%=MOD;
    FOR(j,0,26)if(j!=c)(Y[c]+=Y[j])%=MOD;
    FOR(j,0,26) ans2 += X[j];
  }
  cout << ans << endl;
}