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 MX = 2e5 + 5;
const long long MOD = 1e9 + 7;
long long a[MX], T, n;
long long fact[MX];
long long twoP[MX];
long long inv(long long x, long long m = MOD) {
  return 1 < x ? m - inv(m % x, x) * m / x : 1;
}
long long C(long long x, long long y) {
  return ((fact[x] * (inv(fact[x - y]))) % MOD * inv(fact[y])) % MOD;
}
int main() {
  cin >> n >> T;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  fact[0] = 1ll;
  twoP[0] = 1ll;
  for (int i = 1; i <= n; ++i) {
    fact[i] = (1ll * fact[i - 1] * i) % MOD;
    twoP[i] = (1ll * twoP[i - 1] * 2) % MOD;
  }
  long long val = 1, ans = 0, sum = 0, cnt = 0;
  for (int i = 1; i <= n; ++i) {
    sum += a[i];
    val = (val * 2) % MOD;
    val = (val - C(i - 1, cnt) + MOD) % MOD;
    long long dif = min(T - sum, 1ll * i);
    if (dif < 0) break;
    int k = cnt - dif;
    for (int j = 0; j < k; ++j) {
      val = (val - C(i, cnt) + MOD) % MOD;
      cnt--;
    }
    if (sum + i <= T) {
      cnt++;
      val = (val + C(i, cnt)) % MOD;
      ans += (twoP[n]) % MOD;
      ans %= MOD;
      continue;
    }
    ans += (val * twoP[n - i]) % MOD;
    ans %= MOD;
  }
  ans = (ans * inv(twoP[n])) % MOD;
  cout << ans << "\n";
}
