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 P = 1e9 + 7, inv2 = (P + 1) / 2;
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
const int N = 1e6 + 10;
int n, a[N], fac[N], ifac[N], po[N];
long long T, s[N];
int C(long long n, long long m) {
  if (n < m) return 0;
  return (long long)fac[n] * ifac[m] % P * ifac[n - m] % P;
}
struct {
  int n, m, c;
  void pr() { printf("n=%d,m=%d,c=%d\n", n, m, c); }
} f[N];
int main() {
  fac[0] = ifac[0] = po[0] = 1;
  for (int i = 1; i <= N - 1; ++i) po[i] = po[i - 1] * 2ll % P;
  for (int i = 1; i <= N - 1; ++i) fac[i] = (long long)fac[i - 1] * i % P;
  ifac[N - 1] = inv(fac[N - 1]);
  for (int i = N - 2; i >= 1; --i)
    ifac[i] = (long long)ifac[i + 1] * (i + 1) % P;
  scanf("%d%lld", &n, &T);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i), s[i] = s[i - 1] + a[i];
  int ans = 0, cnt = 0;
  for (int i = 1; i <= n; ++i)
    if (s[i] <= T) {
      int mx = i;
      if (T - s[i] < mx) mx = T - s[i];
      f[++cnt] = {i, mx, po[n - i]};
    }
  int pre = -1, tot = 0;
  for (int i = cnt; i >= 1; --i) {
    if (i != cnt && f[i].m >= 1) {
      if (pre > f[i].m) {
        tot = po[f[i].n];
      } else {
        for (int j = pre + 1; j <= f[i].m; ++j) {
          tot = (tot + C(f[i + 1].n, j)) % P;
        }
        tot = (tot + C(f[i].n, f[i].m)) * (long long)inv2 % P;
      }
    } else {
      tot = 0;
      for (int j = 0; j <= f[i].m; ++j) tot = (tot + C(f[i].n, j)) % P;
    }
    pre = f[i].m;
    ans = (ans + (long long)tot * f[i].c) % P;
  }
  ans = (long long)ans * inv(po[n]) % P;
  if (ans < 0) ans += P;
  printf("%d\n", ans);
}
