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 long long N = 200002, M = 1e9 + 7;
long long g[N], f[N], n, i, now, ans, j, x, p, T;
long long pw(long long x, long long y) {
  long long z = 1;
  for (; y; y >>= 1, x = x * x % M)
    if (y & 1) z = z * x % M;
  return z;
}
long long C(long long n, long long m) {
  return n < m ? 0 : f[n] * g[m] % M * g[n - m] % M;
}
signed main() {
  scanf("%lld%lld", &n, &T);
  f[0] = 1;
  for (i = 1; i < N; i++) f[i] = f[i - 1] * i % M;
  g[N - 1] = pw(f[N - 1], M - 2);
  for (i = N - 2; ~i; i--) g[i] = g[i + 1] * (i + 1) % M;
  for (i = now = 1; i <= n; now = now * 2 - C(i, j), i++) {
    scanf("%lld", &x), p = std::min(T -= x, i);
    if (p < 0) break;
    for (; j < p; j++) now += C(i, j + 1);
    for (; j > p; j--) now -= C(i, j);
    ans += (now %= M) * pw(2, n - i) % M;
  }
  printf("%lld", (ans % M * pw(2, n * (M - 2)) % M + M) % M);
}
