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 Maxn = 200005, p = 1e9 + 7;
int n, x_now, y_now, a[Maxn];
long long T, ans, now = 1, result = 1, sum, fac[Maxn], inv[Maxn];
long long get_inv(long long x) {
  if (x <= 1) return 1;
  return (p - p / x) * get_inv(p % x) % p;
}
long long C(int x, int y) { return fac[x] * inv[y] % p * inv[x - y] % p; }
long long Csum(int x, int y) {
  while (x > x_now) (result = result * 2 + p - C(x_now, y_now)) %= p, x_now++;
  while (y < y_now) (result = result + p - C(x_now, y_now)) %= p, y_now--;
  while (y > y_now) y_now++, (result = result + C(x_now, y_now)) %= p;
  return result;
}
int main() {
  scanf("%d%lld", &n, &T);
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  inv[n] = get_inv(fac[n]);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % p;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    (now *= (p + 1) / 2) %= p;
    sum += a[i];
    if (sum > T) break;
    (ans += now * Csum(i, min((long long)i, T - sum))) %= p;
  }
  printf("%lld", ans);
  return 0;
}
