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 N = 1e6 + 10;
const int M = 1e9 + 7;
const int inf = 1e9 + 7;
const long long INF = 1e18;
int n;
long long T;
long long a[N];
long long b[N];
long long p[N];
long long p2[N];
long long m2 = 500000004;
long long h[N], rh[N];
long long pw(long long a, long long b) {
  long long t = 1;
  while (b) {
    if (b & 1) {
      t = t * a % M;
    }
    b >>= 1;
    a = a * a % M;
  }
  return t;
}
long long C(int n, int k) {
  if (n < k) return 0;
  return h[n] * rh[n - k] % M * rh[k] % M;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> T;
  p2[0] = 1;
  for (int i = 1; i <= n; i++) {
    p2[i] = p2[i - 1] * m2 % M;
  }
  h[0] = rh[0] = 1;
  for (int i = 1; i <= n; i++) {
    h[i] = h[i - 1] * i % M;
  }
  rh[n] = pw(h[n], M - 2);
  for (int i = n; i >= 1; i--) {
    rh[i - 1] = rh[i] * (i) % M;
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = b[i - 1] + a[i];
  }
  b[n + 1] = b[n] + n + 1;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i] > T) break;
    if (b[i + 1] + i + 1 <= T) continue;
    long long k = T - b[i + 1];
    if (k < 0) k = 0;
    while (k <= i && b[i] + k <= T) {
      long long tmp = C(i, k) * i % M;
      tmp = tmp * p2[i] % M;
      if (b[i + 1] + k == T) {
        tmp = tmp * m2 % M;
      }
      ans = (ans + tmp) % M;
      k++;
    }
  }
  cout << ans << "\n";
}
