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;
long long sum[200002], W[200002], N1[200002], E[200002], N2[200002];
long long fast(int x, int y) {
  long long res = 1;
  while (x) {
    if (x & 1) res = res * y % 1000000007;
    y = 1ll * y * y % 1000000007;
    x >>= 1;
  }
  return res;
}
int main() {
  int n, i, t;
  long long T;
  scanf("%d%lld", &n, &T);
  for (i = 1; i <= n; i++) {
    scanf("%d", &t), sum[i] = sum[i - 1] + t;
    if (sum[i] > T) break;
  }
  i--;
  if (i == 0) {
    puts("0");
    return 0;
  }
  W[0] = E[0] = 1;
  for (int j = 1; j <= i; j++) W[j] = W[j - 1] * j % 1000000007;
  for (int j = 0; j <= i; j++) N1[j] = fast(1000000007 - 2, W[j]);
  for (int j = 1; j <= i; j++) E[j] = (E[j - 1] * 2) % 1000000007;
  for (int j = 0; j <= i; j++) N2[j] = fast(1000000007 - 2, E[j]);
  long long ans = 0;
  for (int j = 0; j <= i; j++) {
    long long now = sum[i] + j;
    if (now > T) {
      i--;
      ans = (ans +
             (W[i] * i % 1000000007) * (N1[j - 1] * N1[i - j + 1] % 1000000007 *
                                        N2[i + 1] % 1000000007)) %
            1000000007;
    }
    if (i >= j)
      ans = (ans + (W[i] * i % 1000000007) *
                       (N1[j] * N1[i - j] % 1000000007 * N2[i] % 1000000007)) %
            1000000007;
    else
      break;
  }
  printf("%lld\n", ans);
  return 0;
}
