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 mod = 1e9 + 7;
long long Pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a, res %= mod;
    a *= a;
    a %= mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long x) { return Pow(x, mod - 2); }
long long fac[234567], invfac[234567];
void build() {
  fac[1] = fac[0] = invfac[0] = invfac[1] = 1;
  for (long long i = 2; i < 234567; i++) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= mod;
    invfac[i] = invfac[i - 1] * inv(i);
    invfac[i] %= mod;
  }
}
long long C(long long a, long long b) {
  if (a < 0 || b < 0) return 0;
  if (a < b) return 0;
  return ((fac[a] * invfac[b]) % mod * invfac[a - b]) % mod;
}
long long mul(long long a, long long b) { return (a * b) % mod; }
long long add(long long a, long long b) { return (a + b) % mod; }
long long arr[234567], sum[234567];
int32_t main() {
  ios::sync_with_stdio();
  cin.tie(0);
  build();
  long long n, t;
  cin >> n >> t;
  sum[0] = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
    sum[i] = sum[i - 1] + arr[i];
  }
  long long ans = 0, inv2 = inv(2), totinv2 = inv2;
  sum[n + 1] = t;
  long long CCC = 0, last;
  for (long long i = 1; i <= n; i++) {
    if (sum[i] > t) break;
    long long dif = min(i, t - sum[i]);
    if (i == 1) {
      for (long long j = 0; j <= dif; j++) CCC += C(i, j);
      last = dif;
    } else {
      CCC = add(mul(2, CCC), C(i - 1, last + 1));
      last++;
      while (last > dif) {
        CCC = add(CCC, mod - C(i, last--));
      }
    }
    ans = add(ans, mul(CCC, totinv2));
    totinv2 = mul(totinv2, inv2);
  }
  cout << ans << endl;
}
