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 N = 200005, Mod = 1e9 + 7;
int n, F[N], I[N], Pw[N];
long long T, A[N];
inline int nCr(int a, int b) {
  if (a < 0 || b < 0 || b > a) return 0;
  return (F[a] * 1LL * I[b] % Mod * I[a - b] % Mod);
}
int main() {
  scanf("%d%lld", &n, &T);
  for (int i = 1; i <= n; i++) scanf("%lld", &A[i]);
  for (int i = F[0] = 1; i < N; i++) F[i] = 1LL * F[i - 1] * i % Mod;
  I[0] = I[1] = 1;
  for (int i = 2; i < N; i++) I[i] = Mod - (Mod / i) * 1LL * I[Mod % i] % Mod;
  for (int i = Pw[0] = 1; i < N; i++) Pw[i] = 1LL * Pw[i - 1] * I[2] % Mod;
  for (int i = 1; i < N; i++) I[i] = 1LL * I[i] * I[i - 1] % Mod;
  int Exp = 0, SM = 0, Nw = -1;
  for (int i = 1; i <= n; i++) {
    A[i] += A[i - 1];
    if (A[i] > T) break;
    SM = SM * 2LL % Mod;
    SM = (SM - nCr(i - 1, Nw) + Mod) % Mod;
    if (A[i] + i <= T) {
      Exp = (Exp + 1) % Mod;
      continue;
    }
    int t = min(T - A[i], i * 1LL);
    while (Nw < t) SM = (SM + nCr(i, ++Nw)) % Mod;
    while (Nw > t) SM = (SM - nCr(i, Nw--) + Mod) % Mod;
    Exp = (Exp + 1LL * Pw[i] * SM) % Mod;
  }
  return !printf("%d\n", Exp);
}
