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;
const int N = 2e5 + 5, mod = 1e9 + 7;
int n;
long long T, ans, sum[N], fra[N], inv[N];
long long fpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return res % mod;
}
long long C(int n, int m) {
  if (n < m) {
    return 0;
  }
  return (fra[n] * inv[m] % mod) * inv[n - m] % mod;
}
int main() {
  scanf("%d%lld", &n, &T);
  fra[0] = inv[0] = 1;
  long long x;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    sum[i] = sum[i - 1] + x;
    fra[i] = fra[i - 1] * i % mod;
    inv[i] = inv[i - 1] * fpow(i, mod - 2) % mod;
  }
  long long res = 1, inv2 = fpow(2, mod - 2), tmp = 1;
  for (int i = 1, pos = n; i <= n; i++) {
    res = res * inv2 % mod;
    tmp = (2 * tmp - C(i - 1, pos) + mod) % mod;
    if (T - sum[i] < 0) {
      break;
    }
    while (pos > min(1ll * n, T - sum[i])) {
      tmp = (tmp - C(i, pos) + mod) % mod;
      pos--;
    }
    ans = (ans + res * tmp) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
