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;
void ga(int N, int *A) {
  for (int i(0); i < N; i++) scanf("%d", A + i);
}
long long pw(long long n, long long k) {
  long long r(1);
  while (k) {
    if (k & 1) r *= n, r %= (1000000007);
    n *= n, n %= (1000000007), k >>= 1;
  }
  return r;
}
long long inv(long long a) { return pw(a, (1000000007) - 2); }
long long f[(1 << 20)] = {1, 1}, I[(1 << 20)] = {1, 1}, J[(1 << 20)] = {1, 1};
void pre() {
  for (int k(2); k < (1 << 20); k++)
    f[k] = f[k - 1] * k % (1000000007),
    J[k] =
        -(1000000007) / k * J[(1000000007) % k] % (1000000007) + (1000000007),
    I[k] = I[k - 1] * J[k] % (1000000007);
}
long long C(int N, int K) {
  return N < K ? 0 : (f[N] * I[K] % (1000000007)) * I[N - K] % (1000000007);
}
long long N, T, S[(1 << 20)], O;
int main(void) {
  pre();
  scanf("%lld%lld", &N, &T), S[N] = T + 1;
  for (int i(0); i < N; i++) scanf("%lld", S + i);
  for (int k(1); k < N; k++) S[k] += S[k - 1];
  for (int i(0); i < N; i++) {
    long long l = max(T - S[i + 1], 0ll), r = min(T - S[i] + 1, i + 2ll);
    if (l > r) continue;
    for (int k(l); k < r + 1; k++) {
      if (S[i] + k > T) continue;
      if (S[i + 1] + k <= T) {
        if (S[i + 1] + k + 1 > T)
          O = (O + (i + 1) * C(i + 1, k) % (1000000007) * inv(pw(2, i + 2)) %
                       (1000000007)) %
              (1000000007);
        continue;
      }
      if (S[i] + k > T) continue;
      O = (O + (i + 1) * C(i + 1, k) % (1000000007) * inv(pw(2, i + 1)) %
                   (1000000007)) %
          (1000000007);
    }
  }
  printf("%lld\n", O);
  return 0;
}
