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;
int mod = 1000000007;
long long P(long long a, long long b) {
  if (b == 0) return 1;
  long long x = P(a, b / 2);
  x = (x * x) % mod;
  if (b % 2) x = (x * a) % mod;
  return x;
}
long long inv(long long x) { return P(x, mod - 2); }
long long F[1000000];
long long C(long long n, long long k) {
  if (k > n) return 0;
  if (k == 0) return 1;
  if (k == 1) return n;
  long long A = F[n];
  long long B = inv((F[n - k] * F[k]) % mod);
  return (A * B) % mod;
}
long long t[1000000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  long long T;
  cin >> n >> T;
  F[0] = 1;
  for (int i = 1; i <= n; i++) F[i] = (F[i - 1] * i) % mod;
  long long N = 1;
  long long K = 1;
  long long sum = 2;
  long long cur = 0;
  long long ret = 0;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    cur += t[i];
    if (cur + i <= T) {
      ret = (ret + 1) % mod;
      continue;
    }
    if (cur > T) break;
    long long top = i;
    long long bot = min(i + 0LL, T - cur);
    while (N < top) {
      if (K == N) {
        sum = (sum - C(N, K) + mod) % mod;
        K--;
      }
      sum = (sum * 2 + C(N, K + 1)) % mod;
      N++;
      K++;
    }
    while (K > bot) {
      sum = (sum - C(N, K) + mod) % mod;
      K--;
    }
    ret += (inv(P(2, i)) * sum) % mod;
    ret %= mod;
  }
  cout << ret << endl;
}
