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 = 200005;
const int MOD = 1000000007;
const int INV = (MOD + 1) / 2;
long long f[N], rev[N], fac[N];
long long quickPow(long long a, int b) {
  long long res = 1;
  for (; b; b >>= 1) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
  }
  return res;
}
long long C(int a, int b) {
  if (a < b) {
    return 0;
  }
  return fac[a] * rev[b] % MOD * rev[a - b] % MOD;
}
int main() {
  int n;
  long long T;
  cin >> n >> T;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % MOD;
  }
  rev[n] = quickPow(fac[n], MOD - 2);
  for (int i = n; i >= 1; i--) {
    rev[i - 1] = rev[i] * i % MOD;
  }
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    long long a = T;
    long long b = T - t;
    T = b;
    if (b < 0) {
      break;
    }
    f[i] = 2 * f[i - 1] % MOD;
    if (b >= i) {
      continue;
    }
    long long res = 0;
    for (int j = b; j < i && j <= a; j++) {
      res += C(i - 1, j) * (1 + (j > b)) % MOD;
    }
    f[i] = (f[i] - res % MOD + MOD) % MOD;
  }
  long long ans = 0, mul = 1;
  for (int i = 1; i <= n; i++) {
    mul = mul * INV % MOD;
    ans = (ans + f[i] * mul % MOD) % MOD;
  }
  cout << ans << endl;
  return 0;
}
