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 mxN = 2e5, M = 1e9 + 7, I2 = 5e8 + 4;
int n;
long long T, t[mxN], pI2[mxN + 1], iv[mxN + 1], f1[mxN + 1], f2[mxN + 1], ans;
long long nck(int n, int k) { return f1[n] * f2[k] % M * f2[n - k] % M; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> T;
  for (int i = 0; i < n; ++i) {
    cin >> t[i];
    if (i) t[i] += t[i - 1];
  }
  pI2[0] = 1;
  for (int i = 0; i < n; ++i) pI2[i + 1] = pI2[i] * I2 % M;
  iv[1] = 1;
  for (int i = 2; i <= n; ++i) iv[i] = M - M / i * iv[M % i] % M;
  f1[0] = f2[0] = 1;
  for (int i = 1; i <= n; ++i) {
    f1[i] = f1[i - 1] * i % M;
    f2[i] = f2[i - 1] * iv[i] % M;
  }
  long long rt = 1;
  int nr = 0;
  for (int i = 0; i < n; ++i) {
    rt = (rt - nck(i, nr) * pI2[i + 1] % M + M) % M;
    long long l = t[i], r = t[i] + i + 1;
    if (r > T && l <= T) {
      int tr = r - T - 1;
      while (nr < tr) {
        ++nr;
        rt = (rt + nck(i + 1, nr) * pI2[i + 1]) % M;
      }
      ans = (ans + 1 - rt + M) % M;
    } else if (l <= T)
      ans = (ans + 1) % M;
  }
  cout << ans;
}
