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>
const int MOD = 1e9 + 7;
int ts[200005];
int fact[200005] = {1, 1};
int inv[200005] = {0, 1};
int invfact[200005] = {1, 1};
int choose(int a, int b) {
  if (b < 0 || b > a) return 0;
  return 1LL * fact[a] * invfact[b] % MOD * invfact[a - b] % MOD;
}
int main() {
  for (int i = 2; i <= 200000; i++) {
    inv[i] = 1LL * (MOD - MOD / i) * inv[MOD % i] % MOD;
    fact[i] = 1LL * fact[i - 1] * i % MOD;
    invfact[i] = 1LL * invfact[i - 1] * inv[i] % MOD;
  }
  int N;
  int64_t T;
  scanf("%d %I64d", &N, &T);
  for (int i = 0; i < N; i++) {
    scanf("%d", &ts[i]);
  }
  int ans = 0;
  int den = 1;
  int x = 0;
  int prefix = 1;
  for (int i = 0; i < N; i++) {
    T -= ts[i];
    den = 1LL * den * inv[2] % MOD;
    prefix = (prefix * 2LL + choose(i, ++x)) % MOD;
    if (T < 0) break;
    if (T > i + 1) {
      ans = (ans + 1) % MOD;
    } else {
      while (x > T) {
        prefix = (prefix - choose(i + 1, x--) + MOD) % MOD;
      }
      ans = (ans + 1LL * prefix * den) % MOD;
    }
  }
  printf("%d\n", ans);
  return 0;
}
