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 long long MX = 2e5 + 5;
long long s[MX];
long long a[MX];
long long f[MX];
long long n, m;
const long long D = 1e9 + 7;
long long p;
long long q;
long long pow2(long long x, long long y) {
  if (y == 0) return 1;
  long long ret = pow2(x, y / 2);
  ret *= ret;
  ret %= D;
  if (y % 2) ret *= x;
  return ret % D;
}
long long c(long long x, long long y) {
  return f[x] * pow2(f[y], D - 2) % D * pow2(f[x - y], D - 2) % D;
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  q = pow2(2, n + 1);
  long long i, j;
  f[0] = 1;
  for (i = 1; i <= n; i++) {
    f[i] = f[i - 1] * i % D;
    cin >> a[i];
    s[i] = s[i - 1] + a[i];
  }
  s[n + 1] = m * 2;
  long long w = 0;
  for (i = 1; i <= n; i++) {
    for (j = max(w, m - s[i + 1] + 1); j <= min(m - s[i], i); j++) {
      p = (p + c(i, j) * i % D * pow2(2, n - i) % D) % D;
    }
    for (j = max(w, m - s[i + 1]); j <= min(m - s[i], i); j++)
      p = (p + c(i, j) * i % D * pow2(2, n - i) % D) % D;
  }
  cout << p * pow2(q, D - 2) % D;
  return 0;
}
