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 md = 1e9 + 7;
long long a[300010];
long long fact[300010];
long long step(long long x, long long y) {
  if (y == 0) return 1ll;
  if (y == 1) return x;
  if (y % 2 == 0) {
    long long kik = step(x, y / 2);
    return (kik * kik) % md;
  }
  return (step(x, y - 1) * x) % md;
}
long long C(long long n, long long l, long long r) {
  l = max(0LL, l);
  r = max(0LL, r);
  long long ans = 0;
  for (int k = l; k <= r; k++) {
    ans = (ans +
           fact[n] * step(fact[k], md - 2) % md * step(fact[n - k], md - 2)) %
          md;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  fact[0] = 1;
  for (int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % md;
  }
  long long curt = 0;
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    curt += a[i];
    if (curt > m) break;
    if (i == n) {
      ans = (ans +
             step(step(2LL, i), md - 2) * C(i, 0, min(i, m - curt)) % md * i) %
            md;
    } else if (i == m) {
      ans = (ans + step(step(2LL, i), md - 2) * i) % md;
    } else {
      if (curt + a[i + 1] + i + 1 <= m) continue;
      ans = (ans +
             step(step(2LL, i + 1), md - 2) *
                 C(i, m - curt - a[i + 1], min(i, m - curt)) % md * i +
             step(step(2LL, i + 1), md - 2) *
                 C(i, m + 1 - curt - a[i + 1], min(i, m - curt)) % md * i) %
            md;
    }
  }
  cout << ans;
}
