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 = 2e5 + 10;
const int BASE = 1e9 + 7;
long long lim;
int n;
int a[N];
long long sum[N], fact[N], inv[N], pw2[N];
long long Power(long long a, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * a % BASE;
    a = a * a % BASE;
    p >>= 1;
  }
  return ans;
}
void Add(long long &a, long long b) {
  a = (a + b) % BASE;
  if (a < 0) a += BASE;
}
long long C(int k, int n) {
  assert(k <= n);
  return fact[n] * inv[n - k] % BASE * inv[k] % BASE;
}
long long Compute(int p, int last_problem, int delay_time) {
  bool equal = (sum[p + 1] + delay_time + last_problem == lim);
  return 1LL * p * Power((equal == true ? pw2[p + 1] : pw2[p]), BASE - 2) %
         BASE * C(delay_time, p - 1) % BASE;
}
int main() {
  cin >> n >> lim;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + a[i];
  fact[0] = 1;
  for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i % BASE;
  inv[n] = Power(fact[n], BASE - 2);
  for (int i = n - 1; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1) % BASE;
  pw2[0] = 1;
  for (int i = 1; i <= n; ++i) pw2[i] = pw2[i - 1] * 2 % BASE;
  long long ans = 0;
  for (int last_problem = 0; last_problem < 2; ++last_problem) {
    int p = n;
    for (int delay_time = 0; delay_time <= n; ++delay_time) {
      while (p - 1 >= delay_time && sum[p] + delay_time + last_problem >= lim)
        --p;
      if (p - 1 >= delay_time && sum[p] + delay_time + last_problem < lim) {
        Add(ans, Compute(p, last_problem, delay_time));
      }
    }
  }
  int p = n;
  for (int delay_time = 0; delay_time <= n; ++delay_time) {
    while (p - 1 >= delay_time && sum[p] + delay_time > lim) --p;
    if (p >= delay_time && sum[p] + delay_time == lim) {
      long long ret =
          1LL * p * Power(pw2[p], BASE - 2) % BASE * C(delay_time, p) % BASE;
      Add(ans, ret);
    }
  }
  cout << ans << endl;
  return 0;
}
