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>
#pragma GCC optimize("Ofast")
#pragma optimize("O3")
using namespace std;
long long N = 1e9 + 7, M = 31;
long long i, j, l, r, n, m, t, k, f, ma = 0, y3, mi = 2e9, y, x, z, x2, y2,
                                     dp[300000], a[300000];
long long ans;
long long voz(long long x, long long y) {
  if (y == 0) return 1ll;
  if (y == 1) return x;
  if (y % 2 == 0) {
    long long t = voz(x, y / 2);
    return (t * t) % N;
  }
  if (y % 2 == 1) {
    long long t = voz(x, y / 2);
    return (t * t % N * x) % N;
  }
}
long long C(long long n, long long lk, long long rk) {
  lk = max(0ll, lk);
  rk = max(0ll, rk);
  long long ans = 0;
  for (long long k = lk; k <= rk; k++) {
    ans = (ans + dp[n] * voz(dp[k], N - 2) % N * voz(dp[n - k], N - 2)) % N;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  dp[1] = 1;
  dp[0] = 1;
  for (i = 2; i <= n; i++) {
    dp[i] = (dp[i - 1] * i) % N;
  }
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    t += a[i];
    if (t > m) break;
    if (i == n) {
      l = 0;
      r = min(i, m - t);
      k = n;
      ans = (ans + voz(voz(2ll, i), N - 2) * C(i, l, r) % N * i) % N;
      continue;
    } else if (t == m) {
      l = 0;
      r = 0;
      k = i;
      ans = (ans + voz(voz(2ll, i), N - 2) * i) % N;
      continue;
    } else {
      if (t + a[i + 1] + i + 1 <= m) continue;
      l = m + 1 - t - a[i + 1];
      r = min(i, m - t);
      k = i + 1;
      ans = (ans + voz(voz(2ll, i + 1), N - 2) * C(i, l - 1, r) % N * i +
             voz(voz(2ll, i + 1), N - 2) * C(i, l, r) % N * i) %
            N;
      continue;
    }
  }
  cout << ans;
}
