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 N = 2e5 + 5;
int fact[N], factinv[N], inv[N];
template <typename T, typename T1>
T nCr(T n, T1 r) {
  if (r > n) return 0;
  int ans = (int)((1ll * factinv[r] * factinv[n - r]) % 1000000007);
  ans = (int)((1ll * ans * fact[n]) % 1000000007);
  return ans;
}
template <typename T>
T mod(T x) {
  x %= 1000000007;
  if (x < 0) x += 1000000007;
  return x;
}
template <typename T>
T modinv(T x) {
  x = mod(x);
  if (x == 1) return 1;
  return (
      (int)mod(1ll * (-1000000007 / x) * (inv[1000000007 % x] % 1000000007)));
}
template <typename T, typename T1>
T expo(T a, T1 b) {
  if (b == 0) return 1;
  if (b & 1) return (int)(1ll * a * expo(a, b - 1) % 1000000007);
  a = expo(a, b / 2);
  return (int)(1ll * a * a % 1000000007);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int i, n, k, t;
  int64_t T, S = 0, ans = 0, mi = 1, prvmi = 0, j, sigma = 1;
  cin >> n >> T;
  fact[0] = factinv[0] = 1;
  for (i = 1; i < N; i++) {
    fact[i] = (int)((1ll * fact[i - 1] * i) % 1000000007);
    inv[i] = modinv(i);
    factinv[i] = (int)((1LL * factinv[i - 1] * inv[i]) % 1000000007);
  }
  for (i = 1; i <= n; i++, mi++) {
    cin >> t;
    S += t;
    if (S > T) break;
    mi = (mi < T - S ? mi : T - S);
    if (mi > prvmi)
      for (j = mi; j > prvmi; j--) sigma = mod(sigma + nCr(i, j));
    else
      for (j = prvmi; j > mi; j--) sigma = mod(sigma - nCr(i, j));
    int modinv = expo(2, i);
    modinv = modinv < N ? inv[modinv] : expo(modinv, 1000000007 - 2);
    int64_t Pi = mod(sigma * modinv);
    ans = mod(ans + Pi);
    sigma = mod(sigma * 2);
    sigma = mod(sigma + nCr(i, mi + 1));
    sigma = mod(sigma - nCr(i + 1, mi + 1));
    prvmi = mi;
  }
  cout << ans;
  return 0;
}
