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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, t;
  cin >> n >> t;
  vector<long long> a((size_t)n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> b((size_t)n);
  for (long long i = 0; i < n; i++) {
    b[i] = (i == 0 ? 0 : b[i - 1]) + a[i];
  }
  auto norm = [&](long long x) {
    return (x % 1000000007 + 1000000007) % 1000000007;
  };
  function<long long(long long, long long)> power = [&](long long x,
                                                        long long y) {
    if (y == 0) {
      return (long long)1;
    } else {
      long long z = power(x, y / 2);
      return y % 2 == 0 ? norm(z * z) : norm(norm(z * z) * x);
    }
  };
  auto div = [&](long long x) { return power(x, norm(-2)); };
  vector<long long> factorial((size_t)n + 1, 1);
  for (long long i = 1; i <= n; i++) {
    factorial[i] = norm(factorial[i - 1] * i);
  }
  auto binomial = [&](long long x, long long y) {
    return x < y ? (long long)0
                 : norm(norm(factorial[x] * div(factorial[y])) *
                        div(factorial[x - y]));
  };
  long long o = 0;
  for (long long i = 0, d = div(2), p = 2, m = 1; i < n;
       i++, d = norm(d * div(2)), p = norm(p * 2 + binomial(i, m + 1)), m++) {
    if (b[i] + i + 1 <= t) {
      o = norm(o + 1);
    } else if (b[i] <= t) {
      for (; m > min(t - b[i], i + 1); m--) {
        p = norm(p - binomial(i + 1, m));
      }
      o = norm(o + norm(p * d));
    }
  }
  cout << o << endl;
  return 0;
}
