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 f[200005], v[200005], p[200005], t[200005];
long long ksm(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      ans = ans * a % 1000000007;
    }
    a = a * a % 1000000007;
    b /= 2;
  }
  return ans;
}
long long g(long long x, long long y) {
  return f[x] * v[x - y] % 1000000007 * v[y] % 1000000007;
}
int main() {
  long long i, j, l, r, n, m, k, sum = 0, ans = 0;
  f[0] = p[0] = 1;
  for (i = 1; i <= 200001; i++) {
    f[i] = f[i - 1] * i % 1000000007;
    p[i] = p[i - 1] * 500000004 % 1000000007;
  }
  v[200001] = ksm(f[200001], 1000000007 - 2);
  for (i = 200000; i >= 0; i--) {
    v[i] = v[i + 1] * (i + 1) % 1000000007;
  }
  cin >> n >> m;
  for (i = 1; i <= n; i++) cin >> t[i];
  t[++n] = m + 1;
  for (i = 1; i <= n && sum <= m; i++) {
    if (sum + i + t[i] > m) {
      l = max(0ll, m - sum - t[i] + 1);
      r = min(i - 1, m - sum);
      for (j = l; j <= r; j++) {
        ans = (ans + p[i] * g(i - 1, j) % 1000000007 * (i - 1) % 1000000007) %
              1000000007;
      }
      l = max(0ll, m - sum - t[i]);
      r = min(i - 1, m - sum);
      for (j = l; j <= r; j++) {
        ans = (ans + p[i] * g(i - 1, j) % 1000000007 * (i - 1) % 1000000007) %
              1000000007;
      }
    }
    sum += t[i];
  }
  cout << ans << endl;
  return 0;
}
