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>
const int mod = 1e9 + 7;
using namespace std;
int n, x = 0, y = 0, w = 1, fac[200010], ifac[200010], inv[200010], a[200010];
long long m, sum[200010];
int& qmo(int& x) { return x += (x >> 31) & mod; }
void init(int N) {
  fac[0] = inv[1] = ifac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  for (int i = 2; i <= N; i++)
    inv[i] = mod - 1ll * inv[mod % i] * (mod / i) % mod;
  for (int i = 1; i <= N; i++) ifac[i] = 1ll * ifac[i - 1] * inv[i] % mod;
}
int C(int n, int m) {
  if (n < m || m < 0) return 0;
  return 1ll * fac[n] * ifac[n - m] % mod * ifac[m] % mod;
}
int S(int n, long long m) {
  if (n < 0 || m < 0) return 0;
  m = min(1ll * n, m);
  while (x < n) {
    qmo(qmo(w = w + w - mod) -= C(x, y));
    x++;
  }
  while (x > n) {
    w = 1ll * (w - C(x - 1, y) + mod) * inv[2] % mod;
    x--;
  }
  while (y < m) {
    qmo(w += C(x, y + 1) - mod);
    y++;
  }
  while (y > m) {
    qmo(w -= C(x, y));
    y--;
  }
  return w;
}
int main() {
  scanf("%d%lld", &n, &m);
  init(max(n, 2));
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), sum[i] = sum[i - 1] + a[i];
  int res = 0;
  for (int i = 1, nw = inv[2]; i <= n; i++, nw = 1ll * nw * inv[2] % mod) {
    res = (res + 1ll * nw * S(i, m - sum[i])) % mod;
  }
  printf("%d\n", res);
}
