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 M = 2e5 + 5, P = 1e9 + 7;
int n;
long long T;
int a[M];
long long ans;
long long Pow[M], inv[M];
long long C(int n, int m) { return Pow[n] * inv[m] % P * inv[n - m] % P; }
int main() {
  scanf("%d%lld", &n, &T);
  inv[0] = inv[1] = Pow[0] = Pow[1] = 1;
  for (int i = 2; i <= n + 1; i++) inv[i] = (P - P / i) * inv[P % i] % P;
  for (int i = 2; i <= n + 1; i++) inv[i] = inv[i - 1] * inv[i] % P;
  for (int i = 2; i <= n + 1; i++) Pow[i] = Pow[i - 1] * i % P;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  long long Inv = inv[2], sum = 0, res = 1, ans = 0;
  for (int i = 1, pp = 0; i <= n; i++, Inv = Inv * inv[2] % P) {
    sum += a[i];
    int p = min(T - sum, (long long)i);
    if (p < 0) break;
    res = (res * 2 - C(i - 1, pp)) % P;
    while (pp < p) (res += C(i, ++pp)) %= P;
    while (pp > p) (res -= C(i, pp--)) %= P;
    (ans += res * Inv) %= P;
  }
  printf("%lld\n", (ans + P) % P);
  return 0;
}
