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 = 233333;
const long long mod = 1000000007;
long long Pow(long long a, long long k, long long p) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = 1ll * ans * a % p;
    k >>= 1;
    a = 1ll * a * a % p;
  }
  return ans;
}
long long Inv(long long a, long long p) { return Pow(a, p - 2, p); }
long long inv[N];
void InvList(long long p, int n) {
  inv[0] = 0;
  inv[1] = 1;
  for (int i = 2; i < n; i++) {
    inv[i] = 1ll * inv[p % i] * (p - p / i) % p;
  }
}
long long pow_2[N];
void PowList(long long p, int n) {
  pow_2[0] = 1;
  for (int i = 1; i < n; i++) {
    pow_2[i] = 2 * pow_2[i - 1] % p;
  }
}
long long C[N], pre[N];
long long T, t[N];
long long DP(long long n) {
  InvList(mod, n + 1);
  PowList(mod, n + 1);
  long long r = T, ans = 0;
  pre[0] = 1;
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    r -= t[i];
    if (r >= i) {
      pre[i] = 2 * pre[i - 1] % mod;
      C[i] = 1;
      ans += pre[i] % mod * pow_2[n - i] % mod;
      ans %= mod;
    } else if (r >= 0) {
      long long k = min(r + t[i], i - 1);
      while (k > r) {
        C[k - 1] = C[k] * k % mod * inv[i - k] % mod;
        pre[k - 1] = (pre[k] - C[k]) % mod;
        k--;
      }
      pre[k] = (2 * pre[k] - C[k]) % mod;
      C[k] = C[k] * i % mod * inv[i - k] % mod;
      ans += pre[k] % mod * pow_2[n - i] % mod;
      ans %= mod;
    } else {
      break;
    }
  }
  ans = ans * Inv(pow_2[n], mod) % mod;
  return (ans + mod) % mod;
}
int main() {
  long long n;
  cin >> n >> T;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &t[i]);
  }
  cout << DP(n) << endl;
}
