#include using namespace std; typedef long long ll; typedef vector VI; typedef vector VVI; typedef vector VL; typedef vector> VVL; typedef pair P; typedef tuple tpl; #define ALL(a) (a).begin(),(a).end() #define SORT(c) sort((c).begin(),(c).end()) #define REVERSE(c) reverse((c).begin(),(c).end()) #define EXIST(m,v) (m).find((v)) != (m).end() #define LB(a,x) lower_bound((a).begin(), (a).end(), x) - (a).begin() #define UB(a,x) upper_bound((a).begin(), (a).end(), x) - (a).begin() #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define REP(i,n) FOR(i,0,n) #define RFOR(i,a,b) for(int i=(a)-1;i>=(b);--i) #define RREP(i,n) RFOR(i,n,0) #define en "\n" constexpr double EPS = 1e-9; constexpr double PI = 3.1415926535897932; constexpr int INF = 2147483647; constexpr long long LINF = 1LL<<60; constexpr long long MOD = 1000000007; // 998244353; template inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; } template inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; } struct mint { long long x; mint(long long x=0):x((x%MOD+MOD)%MOD){} mint operator-() const { return mint(-x);} mint& operator+=(const mint a) { if ((x += a.x) >= MOD) x -= MOD; return *this; } mint& operator-=(const mint a) { if ((x += MOD-a.x) >= MOD) x -= MOD; return *this; } mint& operator*=(const mint a) { (x *= a.x) %= MOD; if(x < 0) x += MOD; return *this; } mint operator+(const mint a) const { mint res(*this); return res+=a; } mint operator-(const mint a) const { mint res(*this); return res-=a; } mint operator*(const mint a) const { mint res(*this); return res*=a; } mint pow(long long t) const { if (!t) return 1; mint a = pow(t>>1); a *= a; if (t&1) a *= *this; return a; } // for prime MOD mint inv() const { return pow(MOD-2); } mint& operator/=(const mint a) { return (*this) *= a.inv(); } mint operator/(const mint a) const { mint res(*this); return res/=a; } }; void Main(){ string S; cin >> S; int N = S.size(); VL p(N,1); REP(i,N-1) (p[i+1] *= p[i]*2) %= MOD; mint sub[26], dp[N+1]; REP(i,N){ int x = S[i]-'a'; dp[i+1] = dp[i]*2 + p[i] - sub[x]; sub[x] += p[i]; } int ans = dp[N].x; cout << ans << en; return; } int main(void){ cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);cout<>t; REP(_,t) Main(); return 0; }