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 + 50;
const int mod = 1e9 + 7;
long long fac[maxn], inv[maxn];
long long a[maxn];
long long qpow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
long long C(int n, int k) { return fac[n] * inv[n - k] % mod * inv[k] % mod; }
long long now = 1, nn, kk;
long long cal(int n, int k) {
  while (nn < n) {
    now = (now + now + mod) % mod;
    now = (now - C(nn, kk) + mod) % mod;
    nn++;
    kk++;
  }
  while (kk > k) {
    kk--;
    now = (now + C(nn, kk) + mod) % mod;
  }
  while (kk < k) {
    now = (now - C(nn, kk) + mod) % mod;
    kk++;
  }
  return (now + mod) % mod;
}
int main() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= maxn - 4; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = qpow(fac[i], mod - 2);
  }
  long long n, t;
  cin >> n >> t;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = (a[i - 1] + a[i]);
  }
  long long ans = 0, di = 1;
  for (int i = 1; i <= n; i++) {
    di = di * inv[2] % mod;
    if (a[i] > t)
      break;
    else if (a[i] + i <= t)
      ans = (ans + 1) % mod;
    else {
      ans = ((ans + (cal(i, a[i] + i - t) * di) % mod) % mod + mod) % mod;
    }
  }
  cout << ans << endl;
  return 0;
}
