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 mod = 1e9 + 7, N = 2e5 + 10, M = sqrt(2e5) + 10;
int power(int a, int b, int ans = 1) {
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
int f[N], rf[N];
int C(int n, int m) {
  if (m < 0 || m > n) return 0;
  return 1ll * f[n] * rf[m] % mod * rf[n - m] % mod;
}
int main() {
  for (int i = f[0] = 1; i < N; i++) f[i] = 1ll * f[i - 1] * i % mod;
  rf[N - 1] = power(f[N - 1], mod - 2);
  for (int i = N - 1; i; i--) rf[i - 1] = 1ll * rf[i] * i % mod;
  int n;
  long long T, L = 0;
  cin >> n >> T;
  int ans = 0, crr = 1, pos = 0;
  for (int i = 1; i <= n; i++) {
    int d = 1;
    scanf("%d", &d);
    L += d;
    if (T < L) break;
    crr = (2ll * crr + mod - C(i - 1, pos)) % mod;
    int f = min(1ll * i, T - L);
    for (; pos < f; crr = (crr + C(i, ++pos)) % mod)
      ;
    for (; pos > f; crr = (crr + mod - C(i, pos--)) % mod)
      ;
    ans = (ans + 1ll * crr * power(power(2, i), mod - 2)) % mod;
  }
  cout << ans << endl;
  return 0;
}
