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 N = 1e6 + 5;
const int ha = 1e9 + 7;
long long A[N], f[N], g[N], h[N], B = 440;
struct pp {
  int l, r;
} s[N];
int tot;
bool cmp(pp w, pp v) {
  if (w.l / B == v.l / B) return w.r < v.r;
  return w.l < v.l;
}
long long inv(long long a) {
  long long r = 1, b = ha - 2;
  while (b) {
    if (b & 1) r = r * a % ha;
    a = a * a % ha;
    b >>= 1;
  }
  return r;
}
int C(int n, int m) {
  if (n < m) return 0;
  return (long long)f[n] * g[n - m] % ha * g[m] % ha;
}
int main() {
  f[0] = g[0] = h[0] = 1;
  for (int i = 1; i < N; ++i) {
    f[i] = f[i - 1] * i % ha;
    g[i] = inv(f[i]);
    h[i] = h[i - 1] * 2 % ha;
  }
  long long n, T;
  cin >> n >> T;
  long long IP2 = inv(h[n]);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> A[i];
    A[i] += A[i - 1];
  }
  for (int i = 0, r = n; i <= n && r >= i;) {
    while (A[r] + i > T && r >= i) --r, IP2 = IP2 * 2 % ha;
    if (i > 0 && r >= i - 1) ans += C(r, i - 1) * IP2 % ha * g[2] % ha * r % ha;
    while (A[r] + i <= T && r >= i) ans += C(r, i++) * IP2 % ha * r % ha;
  }
  cout << ans % ha << endl;
  return 0;
}
