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 N;
long long T;
const long long MOD = 1e9 + 7;
long long fact[200005], pows[200005];
long long mult(long long a, long long b) { return a * b % MOD; }
long long add(long long a, long long b) { return (a + b) % MOD; }
long long qikpow(long long a, long long b) {
  if (!b) {
    return 1;
  }
  long long ret = qikpow(a, b / 2);
  ret = mult(ret, ret);
  if (b & 1) {
    return mult(ret, a);
  }
  return ret;
}
long long choose(long long a, long long b) {
  return mult(fact[a], qikpow(mult(fact[b], fact[a - b]), MOD - 2));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> T;
  long long mn = 0, mx = 0;
  long long ans = 0, v = 1, k = 0;
  fact[0] = 1;
  pows[0] = 1;
  for (int i = 1; i <= N; i++) {
    fact[i] = mult(fact[i - 1], i);
    pows[i] = mult(pows[i - 1], 2);
    int n;
    cin >> n;
    mn += n;
    mx += n + 1;
    if (mn > T) {
      break;
    } else if (mx > T) {
      v = mult(v, 2);
      v = add(v, MOD - choose(i - 1, k));
      mx--;
      while (mx > T) {
        v = add(v, MOD - choose(i, k));
        k--;
        mx--;
      }
      ans = add(ans, mult(v, qikpow(pows[i], MOD - 2)));
    } else {
      k++;
      v = mult(v, 2);
      ans++;
    }
  }
  cout << ans;
}
