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>
int d_, x_, y_;
void gcd_(int a, int b) {
  if (b == 0)
    d_ = a, x_ = 1, y_ = 0;
  else {
    int tmp;
    gcd_(b, a % b);
    tmp = x_ - a / b * y_, x_ = y_, y_ = tmp;
  }
}
int inv(int a) {
  gcd_(a, 1000000007);
  return x_;
}
int ff[200000 + 1], gg[200000 + 1];
void init() {
  int i, f;
  f = 1;
  for (i = 0; i <= 200000; i++) {
    gg[i] = inv(ff[i] = f);
    f = (long long)f * (i + 1) % 1000000007;
  }
}
int choose(int n, int k) {
  return (long long)ff[n] * gg[k] % 1000000007 * gg[n - k] % 1000000007;
}
int main() {
  int n, k, ans;
  long long s, x, p;
  init();
  scanf("%d%lld", &n, &s);
  ans = 0, x = 1, p = 1;
  for (k = 1; k <= n; k++) {
    int t;
    long long s_;
    scanf("%d", &t);
    if ((s_ = s - t) < 0) break;
    x = (x * 2 - (k - 1 >= s ? choose(k - 1, s) : 0)) % 1000000007;
    if (s_ < k) {
      if (s > k) s = k;
      while (s > s_) x = (x - choose(k, s--)) % 1000000007;
    } else
      s = s_;
    ans =
        (ans + x * (p = p * ((1000000007 + 1) / 2) % 1000000007)) % 1000000007;
  }
  if (ans < 0) ans += 1000000007;
  printf("%d\n", ans);
  return 0;
}
