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 MOD = int(1e9) + 7;
const long long int MOD64 = (long long int)(1e18) + 7;
const int INF = 0x3fffffff;
const long long int INF64 = 0x3fffffffffffffff;
const int N = 2e5 + 10;
long long int f[N], finv[N];
long long int pow2[N], pow2inv[N];
long long int powmod(long long int a, long long int b, long long int P) {
  long long int t = 1;
  for (; b; b >>= 1, a = a * a % P) {
    if (b & 1) t = t * a % P;
  }
  return t;
}
long long int inv(long long int x, long long int p) {
  return powmod(x, p - 2, p);
}
void init() {
  f[0] = pow2[0] = 1;
  for (int i = int(1); i < int(N); i++) {
    f[i] = f[i - 1] * i % MOD;
    pow2[i] = pow2[i - 1] * 2 % MOD;
  }
  finv[N - 1] = inv(f[N - 1], MOD);
  pow2inv[N - 1] = inv(pow2[N - 1], MOD);
  for (int i = N - 1; i; i--) {
    finv[i - 1] = finv[i] * i % MOD;
    pow2inv[i - 1] = pow2inv[i] * 2 % MOD;
  }
}
long long int comb(int n, int m) {
  if (m < 0 || m > n) return 0;
  return f[n] * finv[n - m] % MOD * finv[m] % MOD;
}
int t[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  long long int T;
  cin >> n >> T;
  for (int i = int(0); i < int(n); i++) cin >> t[i + 1];
  init();
  long long int sum = 0;
  long long int r = 0;
  long long int ans = 0;
  long long int S = 0;
  for (int i = int(1); i < int(n + 1); i++) {
    sum += t[i];
    if (sum + i <= T)
      ans++;
    else if (sum > T)
      break;
    else {
      if (S == 0) {
        for (int j = 0; j + sum <= T; j++) {
          S += comb(i, j);
          r = j;
        }
        S %= MOD;
      } else {
        S = (S - comb(i - 1, r) + MOD) % MOD;
        while (sum + r > T) {
          S = (S - comb(i, r) + MOD) % MOD;
          r--;
        }
      }
      ans += S * pow2inv[i] % MOD;
      S = S * 2 % MOD;
    }
  }
  cout << ans % MOD << endl;
  return 0;
}
