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;
long long t;
int n, ans;
int a[200005], half[200005], f[200005], inv[200005];
long long psum[200005];
int add(int x, int y) {
  x += y;
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
  return x;
}
int mul(int x, int y) { return (long long)x * y % 1000000007; }
int C(int x, int y) { return mul(f[x], mul(inv[x - y], inv[y])); }
int fe(int x, int y) {
  if (y == 0) return 1;
  if (y & 1) return mul(x, fe(x, y - 1));
  int res = fe(x, y >> 1);
  return mul(res, res);
}
void pre() {
  for (int i = 1; i <= n; i++) psum[i] = psum[i - 1] + a[i];
  half[0] = 1;
  half[1] = fe(2, 1000000007 - 2);
  for (int i = 2; i <= n; i++) half[i] = mul(half[i - 1], half[1]);
  f[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = mul(f[i - 1], i);
    inv[i] = fe(f[i], 1000000007 - 2);
  }
}
int main() {
  scanf("%d %lld", &n, &t);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  pre();
  for (int i = 0; i <= n; i++) {
    int bas = 1, son = n;
    while (bas <= son) {
      if (psum[((bas + son) / 2)] + i > t)
        son = ((bas + son) / 2) - 1;
      else
        bas = ((bas + son) / 2) + 1;
    }
    if (son < i) continue;
    ans = add(ans, mul(mul(C(son, i), half[son]), son));
  }
  for (int i = 1; i <= n; i++) {
    int bas = 1, son = n;
    while (bas <= son) {
      if (psum[((bas + son) / 2)] + i > t)
        son = ((bas + son) / 2) - 1;
      else
        bas = ((bas + son) / 2) + 1;
    }
    if (son < i - 1 || psum[bas] + i - 1 != t) continue;
    ans = add(ans, mul(mul(C(son, i - 1), half[bas]), son));
  }
  printf("%d", ans);
}
