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 int mod = 1e9 + 7;
long long int n, i, j, ans, T, a[200005], sum[200005];
vector<long long int> f(200005, 0), invf(200005, 0);
long long int mul(long long int a, long long int b) {
  return (((a % mod) * (b % mod)) % mod);
}
long long int bpow(long long int a, long long int x) {
  long long int res = 1;
  while (x > 0) {
    if (x & 1) res = mul(res, a);
    x = x >> 1;
    a = mul(a, a);
  }
  return (res % mod);
}
long long int inv(long long int a) { return bpow(a, mod - 2); }
long long int c(long long int k, long long int n) {
  return mul(mul(f[n], invf[n - k]), invf[k]);
}
int main() {
  cin >> n >> T;
  long long int res = 1;
  sum[0] = 0;
  f[0] = 1;
  invf[0] = 1;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    sum[i] = sum[i - 1] + a[i];
    res = (res * i) % mod;
    f[i] = res;
    invf[i] = inv(res);
  }
  sum[n + 1] = sum[n] + 1e9;
  if (sum[n] <= T) {
    i = n;
  } else {
    for (i = 1; i <= n; i++) {
      if ((T - sum[i]) < 0) break;
    }
    i--;
  }
  ans = 0;
  long long int cnt = 0;
  while (T - sum[i + 1] <= i) {
    for (j = max(0ll, T - sum[i + 1]); j <= min(T - sum[i], i); j++) {
      long long int res = c(j, i);
      if (j == (T - sum[i + 1])) res = (res * invf[2]) % mod;
      ans = (ans + (((res * i) % mod) * (inv(bpow(2, i)) % mod)) % mod) % mod;
    }
    i--;
  }
  cout << ans;
}
