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;
int n, t[200005];
long long T, fac[200005], _inv[200005], inv[200005], h[200005], g[200005],
    s[200005], ans;
long long C(int n, int m) {
  if (m > n) return 1;
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
int main() {
  scanf("%d%lld", &n, &T);
  fac[0] = 1;
  for (int i = 1; i < 200005; i++) fac[i] = fac[i - 1] * i % 1000000007;
  _inv[0] = _inv[1] = 1;
  for (int i = 2; i < 200005; i++)
    _inv[i] =
        ((1000000007 - 1000000007 / i) * _inv[1000000007 % i]) % 1000000007;
  inv[0] = inv[1] = 1;
  for (int i = 2; i < 200005; i++) inv[i] = inv[i - 1] * _inv[i] % 1000000007;
  h[0] = 1;
  for (int i = 1; i < 200005; i++) h[i] = h[i - 1] * _inv[2] % 1000000007;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t[i]);
    s[i] = s[i - 1] + t[i];
    g[i] = g[i - 1] + t[i] + 1;
  }
  g[n + 1] = T + 1;
  for (int i = 1; i <= n; i++) {
    if (g[i] > T && T >= s[i - 1]) {
      for (int j = max(0ll, T - s[i] + 1); j <= i && s[i - 1] + j - 1 <= T;
           j++) {
        if (s[i - 1] + j > T) {
          ans = (ans + (long long)(i - 1) * h[i] % 1000000007 *
                           C(i - 1, j - 1) % 1000000007) %
                1000000007;
        } else {
          ans = (ans + (long long)(i - 1) * h[i] % 1000000007 * C(i, j) %
                           1000000007) %
                1000000007;
        }
      }
    }
  }
  for (int i = 0; i <= T - s[n]; i++)
    ans = (ans + (long long)n * h[n] % 1000000007 * C(n, i) % 1000000007) %
          1000000007;
  printf("%lld\n", ans);
}
