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;
long long a[1000001], fact[1000001], inv[1000001];
long long comb(long long k, long long n) {
  if (k == 0 || k == n) return 1;
  long long rez = fact[n];
  rez = rez * inv[n - k];
  rez %= 1000000007;
  rez = rez * inv[k];
  rez %= 1000000007;
  return rez;
}
long long mypow(long long x, long long p) {
  long long rez = 1;
  while (p > 0) {
    if (p % 2 == 1) {
      rez = (rez * x) % 1000000007;
    }
    x = (x * x) % 1000000007;
    p /= 2;
  }
  return rez;
}
int main() {
  long long n, m, i, sum, it, it2, ans, sus, jos;
  fact[0] = 1;
  inv[0] = 1;
  for (i = 1; i <= 500000; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
    inv[i] = mypow(fact[i], 1000000007 - 2);
  }
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    a[i]++;
  }
  sum = 0;
  it = 0;
  while (sum + a[it + 1] <= m) {
    it++;
    sum += a[it];
  }
  it2 = it;
  ans = 0;
  m++;
  for (i = 0; i <= it2; i++) {
    while (it2 < n && sum + a[it2 + 1] <= m) {
      it2++;
      sum += a[it2];
    }
    if (sum < m)
      it = it2;
    else
      it = it2 - 1;
    sus = comb(i, it);
    jos = mypow(2, it2);
    ans = (ans + it * (sus * mypow(jos, 1000000007 - 2) % 1000000007)) %
          1000000007;
    sum--;
  }
  cout << ans << '\n';
  return 0;
}
