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;
int n, m;
int inv[200005], fac[200005], pow2[200005], nny[200005];
long long t, x;
long long C(long long a, long long b) {
  return 1ll * fac[a] * inv[a - b] % 1000000007 * inv[b] % 1000000007;
}
long long ny(long long x) {
  return x == 1
             ? 1
             : ny(1000000007 % x) * (1000000007 - 1000000007 / x) % 1000000007;
}
int now;
long long csum(int n, int l, int r) {
  long long ans = 0;
  for (int i = l; i <= r; i++) ans = (ans + C(n, i)) % 1000000007;
  return ans;
}
void init() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i < 200005; i++) {
    long long temp = 1ll * fac[i - 1] * i % 1000000007;
    fac[i] = temp;
    inv[i] = ny(fac[i]);
  }
}
int main() {
  int T;
  pow2[0] = 1;
  for (int i = (1); i <= (200005 - 1); i++)
    pow2[i] = 1ll * pow2[i - 1] * 2 % 1000000007;
  init();
  cin >> n >> t;
  long long sum = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    if (sum <= t && sum + x + i > t) {
      long long l = max(0ll, t - x - sum + 1), r = min(t - sum, 0ll + i - 1);
      if (r >= l)
        ans += 1ll * (i - 1) * (csum(i - 1, l, r)) % 1000000007 * pow2[n - i] %
               1000000007;
      ans %= 1000000007;
      l = max(0ll, t - x - sum), r = min(t - sum, 0ll + i - 1);
      if (r >= l)
        ans += 1ll * (i - 1) * (csum(i - 1, l, r)) % 1000000007 * pow2[n - i] %
               1000000007;
      ans %= 1000000007;
    }
    sum += x;
  }
  if (sum <= t) {
    long long l = 0, r = min(t - sum, 0ll + n);
    if (r >= l) ans += 1ll * n * (csum(n, l, r)) % 1000000007;
    ans %= 1000000007;
  }
  printf("%lld\n", ans * ny(pow2[n]) % 1000000007);
  return 0;
}
