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;
inline void du(long long &d) {
  char t = getchar();
  bool mark = false;
  for (; t < '0' || t > '9'; t = getchar())
    if (t == '-') mark = !mark;
  for (d = 0; t >= '0' && t <= '9'; t = getchar()) d = d * 10 + t - '0';
  if (mark) d = -d;
}
const int maxn = 2e5 + 100;
const long long Mod = 1e9 + 7;
long long a[maxn];
long long sum[maxn];
long long fact[maxn], inv[maxn];
long long Montgemery(long long a, long long b, long long c) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % c;
    a = a * a % c;
    b >>= 1;
  }
  return ans;
}
long long C(long long n, long long m) {
  return (m > n) ? 0 : fact[n] * inv[m] % Mod * inv[n - m] % Mod;
}
int main() {
  long long n, T;
  du(n);
  du(T);
  fact[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) {
    du(a[i]);
    sum[i] = sum[i - 1] + a[i];
    fact[i] = fact[i - 1] * i % Mod;
    inv[i] = inv[i - 1] * Montgemery(i, Mod - 2, Mod) % Mod;
  }
  long long ans = 0, presum = 1;
  long long M1 = 0, M2;
  long long i, j;
  for (i = 1; i <= n; i++) {
    if (sum[i] > T) break;
    ans = ans * 2 % Mod;
    M2 = min(T - sum[i], i);
    if (sum[i] + i <= T) {
      presum = presum * 2 % Mod;
      ans = (ans + presum) % Mod;
      M1 = M2;
      continue;
    }
    for (j = M1; j >= M2; j--) presum = (presum + Mod - C(i - 1, j)) % Mod;
    presum = presum * 2 % Mod;
    presum = (presum + C(i - 1, M2)) % Mod;
    ans = (ans + presum) % Mod;
    M1 = M2;
  }
  long long temp = Montgemery(2, i - 1, Mod);
  ans = ans * Montgemery(temp, Mod - 2, Mod) % Mod;
  cout << ans;
}
