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 inv[200001], fact[200001], inv_fact[200001];
void precomp() {
  inv[1] = 1;
  for (long long ctr1 = 2; ctr1 < 200001; ++ctr1)
    inv[ctr1] = (1000000007 -
                 (1000000007 / ctr1) * inv[1000000007 % ctr1] % 1000000007) %
                1000000007;
  fact[0] = inv_fact[0] = 1;
  for (long long ctr1 = 1; ctr1 < 200001; ++ctr1) {
    fact[ctr1] = fact[ctr1 - 1] * ctr1 % 1000000007;
    inv_fact[ctr1] = 1LL * inv_fact[ctr1 - 1] * inv[ctr1] % 1000000007;
  }
}
int comb(int n, int k) {
  if (k < 0 || k > n) return 0;
  return (1LL * fact[n] * inv_fact[k] % 1000000007) * inv_fact[n - k] %
         1000000007;
}
int pn = 0, pk = 0;
long long psum = 1;
int prefix_sum(int n, int k) {
  while (pn < n) psum = (psum * 2 + comb(pn++, (pk++) + 1)) % 1000000007;
  while (pk > k) psum = (psum - comb(pn, pk--) + 1000000007) % 1000000007;
  return psum;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  precomp();
  int n, a;
  long long t, tot = 0;
  cin >> n >> t;
  long long rezp = 0, rezq = 1;
  for (long long ctr1 = 0; ctr1 < n; ++ctr1) {
    cin >> a;
    tot += a;
    rezp = (rezp * 2) % 1000000007;
    rezq = (rezq * inv[2]) % 1000000007;
    rezp = (rezp + prefix_sum(ctr1 + 1, min(ctr1 + 1, max(t - tot, -1LL)))) %
           1000000007;
  }
  cout << (rezp * rezq) % 1000000007 << '\n';
  return 0;
}
