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 jc[200001];
long long Jc[200001];
long long w[200001];
int t[200001];
int n, i, j, k, l;
long long T, s, ans;
long long qpower(long long a, int b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
long long C(int n, int m) {
  return jc[n] * Jc[m] % 1000000007 * Jc[n - m] % 1000000007;
}
int main() {
  scanf("%d%I64d", &n, &T);
  for (i = 1; i <= n; i++) scanf("%d", &t[i]);
  jc[0] = 1;
  jc[1] = 1;
  Jc[0] = 1;
  Jc[1] = 1;
  w[1] = 1;
  for (i = 2; i <= n; i++) {
    w[i] = 1000000007 -
           (long long)(1000000007 / i) * w[1000000007 % i] % 1000000007;
    jc[i] = jc[i - 1] * i % 1000000007;
    Jc[i] = Jc[i - 1] * w[i] % 1000000007;
  }
  s = 1;
  for (i = 1; i <= n; i++) {
    s = s * 2 % 1000000007;
    if (T <= i - 1) s -= C(i - 1, T);
    T -= t[i];
    if (T < 0) break;
    if (T < i) {
      for (j = (T + t[i] < i ? T + t[i] : i); j >= T + 1; j--)
        s = (s - C(i, j)) % 1000000007;
      if (s < 0) s += 1000000007;
      ans = (ans + s * qpower(500000004, i) % 1000000007) % 1000000007;
    } else
      ++ans;
  }
  printf("%I64d\n", ans);
}
