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 maxn = 2e5 + 10;
const long long mod = 1e9 + 7;
const long long inv2 = 5e8 + 4;
long long fac[maxn], invf[maxn];
long long fpow(long long a, long long k) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    k >>= 1;
  }
  return ans;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i < maxn; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  invf[maxn - 1] = fpow(fac[maxn - 1], mod - 2);
  for (int i = maxn - 2; i >= 0; i--) {
    invf[i] = 1ll * invf[i + 1] * (i + 1) % mod;
  }
}
long long C(long long n, long long m) {
  if (m < 0 || n < m) return 0;
  return fac[n] * invf[n - m] % mod * invf[m] % mod;
}
long long inv(long long k) { return fpow(k % mod, mod - 2); }
long long a[maxn], last = 0, now, s, sum, two;
int main() {
  std::ios::sync_with_stdio(false);
  init();
  long long n, T;
  cin >> n >> T;
  for (int i = 1; i <= n; i++) cin >> a[i];
  s = 1;
  sum = 0;
  two = 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    sum = sum + a[i];
    two = two * inv2 % mod;
    if (sum > T) break;
    now = min(T - sum, i * 1ll);
    if (now == i)
      s = 2ll * s % mod;
    else {
      long long tmp = s;
      for (int j = now; j <= last; j++) tmp = (tmp - C(i - 1, j) + mod) % mod;
      s = (C(i - 1, now) + 2ll * tmp % mod) % mod;
    }
    ans = (ans + s * two % mod) % mod;
    last = now;
  }
  cout << ans << "\n";
}
