Today Adilbek is taking his probability theory test. Unfortunately, when Adilbek arrived at the university, there had already been a long queue of students wanting to take the same test. Adilbek has estimated that he will be able to start the test only T seconds after coming. 

Fortunately, Adilbek can spend time without revising any boring theorems or formulas. He has an app on this smartphone which contains n Japanese crosswords to solve. Adilbek has decided to solve them all one by one in the order they are listed in the app, without skipping any crossword. For each crossword, a number t_i is given that represents the time it takes an average crossword expert to solve this crossword (the time is given in seconds).

Adilbek is a true crossword expert, but, unfortunately, he is sometimes unlucky in choosing the way to solve the crossword. So, it takes him either t_i seconds or t_i + 1 seconds to solve the i-th crossword, equiprobably (with probability 1/2 he solves the crossword in exactly t_i seconds, and with probability 1/2 he has to spend an additional second to finish the crossword). All these events are independent.

After T seconds pass (or after solving the last crossword, if he manages to do it in less than T seconds), Adilbek closes the app (if he finishes some crossword at the same moment, that crossword is considered solved; otherwise Adilbek does not finish solving the current crossword at all). He thinks it would be an interesting probability theory problem to calculate E — the expected number of crosswords he will be able to solve completely. Can you calculate it? 

Recall that the expected value of a discrete random variable is the probability-weighted average of all possible values — in this problem it means that the expected value of the number of solved crosswords can be calculated as E = ∑ _{i = 0}^{n} i p_i, where p_i is the probability that Adilbek will solve exactly i crosswords. 

We can represent E as rational fraction P/Q with Q > 0. To give the answer, you should print P ⋅ Q^{-1} mod (10^9 + 7).

Input

The first line contains two integers n and T (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ T ≤ 2 ⋅ 10^{14}) — the number of crosswords and the time Adilbek has to spend, respectively.

The second line contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ 10^9), where t_i is the time it takes a crossword expert to solve the i-th crossword.

Note that Adilbek solves the crosswords in the order they are given in the input without skipping any of them.

Output

Print one integer — the expected value of the number of crosswords Adilbek solves in T seconds, expressed in the form of P ⋅ Q^{-1} mod (10^9 + 7).

Examples

Input


3 5
2 2 2


Output


750000007


Input


3 5
2 1 2


Output


125000003

Note

The answer for the first sample is equal to 14/8.

The answer for the second sample is equal to 17/8.
#include <bits/stdc++.h>
using namespace std;
const int Mod = 1000000007;
using ll = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VLL = vector<VL>;
int N;
ll T;
VL t;
VI p;
VI fact;
void ReadData();
void Solve();
void CalcP();
int Comb(int N, int K);
int Fact(int x);
int InvMod(int x);
int Pow(int a, int p);
int main() {
  ReadData();
  Solve();
  return 0;
}
void ReadData() {
  cin >> N >> T;
  t = VL(N + 1);
  p = VI(N + 1);
  for (int i = 1; i <= N; ++i) cin >> t[i];
}
void Solve() {
  CalcP();
  int ans{0};
  for (int i = 1; i < N; ++i) {
    ans = (ans + p[i]) % Mod;
  }
  ans = (ans + p[N]) % Mod;
  cout << ans << '\n';
}
void CalcP() {
  ll sum{0}, m;
  for (int i = 1; i <= N; ++i) {
    sum += t[i];
    if (sum > T) break;
    if (sum + i <= T) {
      p[i] = Pow(2, i);
      m = i;
      continue;
    }
    if (i == 1) {
      p[i] = (1LL * InvMod(2) * Pow(2, i)) % Mod;
      m = 0;
      continue;
    }
    int prob = p[i - 1];
    while (sum + m > T) {
      prob = (prob - Comb(i - 1, m) + Mod) % Mod;
      --m;
    }
    prob = (2LL * prob) % Mod;
    prob = (prob - Comb(i - 1, 0) + Mod) % Mod;
    prob = (prob - Comb(i - 1, m) + Mod) % Mod;
    prob = (prob + Comb(i, 0)) % Mod;
    p[i] = prob;
  }
  for (int i = 1; i <= N; ++i) {
    p[i] = (1LL * p[i] * InvMod(Pow(2, i))) % Mod;
  }
}
int Comb(int N, int K) {
  if (K == 0 || N == K) return 1;
  int res = (1LL * Fact(N) * InvMod(Fact(N - K))) % Mod;
  res = (1LL * res * InvMod(Fact(K))) % Mod;
  return res;
}
int Fact(int x) {
  if (static_cast<int>(fact.size()) < x + 1) fact.resize(x + 1);
  if (x == 0) return 1;
  if (fact[x]) return fact[x];
  return fact[x] = (1LL * Fact(x - 1) * x) % Mod;
}
int InvMod(int x) { return Pow(x, Mod - 2); }
int Pow(int a, int p) {
  int res{1};
  while (p) {
    if (p & 1) res = (1LL * res * a) % Mod;
    a = (1LL * a * a) % Mod;
    p >>= 1;
  }
  return res;
}
