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 long long mod = 1e9 + 7;
long long mul(long long a, long long b) { return (a * b) % mod; }
long long bin_pow(long long a, long long b) {
  long long ans = 1;
  long long tmp = a;
  while (b) {
    if (b & 1) {
      ans = mul(ans, tmp);
    }
    tmp = mul(tmp, tmp);
    b >>= 1;
  }
  return ans;
}
long long inv(long long a) { return bin_pow(a, mod - 2); }
const int maxn = 3e5;
long long factorial[maxn];
long long c(int n, int k) {
  if (k > n) return 0;
  return mul(factorial[n], mul(inv(factorial[n - k]), inv(factorial[k])));
}
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) {
  long long ans = (a - b) % mod;
  if (ans < 0) ans += mod;
  return ans;
}
void solve() {
  factorial[0] = 1;
  for (int i = 1; i < maxn; i++) {
    factorial[i] = mul(factorial[i - 1], i);
  }
  int n;
  cin >> n;
  long long T;
  cin >> T;
  long long sum_t = 0;
  long long sum_c = 1;
  long long old_sum_len = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int ti;
    cin >> ti;
    sum_t += ti;
    sum_c = add(mul(2, sum_c), c(i - 1, old_sum_len + 1));
    old_sum_len++;
    long long exp_sum_len = min((long long)i, T - sum_t);
    if (exp_sum_len < 0) exp_sum_len = -1;
    while (old_sum_len != exp_sum_len) {
      sum_c = sub(sum_c, c(i, old_sum_len));
      old_sum_len--;
    }
    long long tmp = mul(sum_c, inv(bin_pow(2, i)));
    ans = add(ans, tmp);
  }
  cout << ans;
}
int main() {
  solve();
  return 0;
}
