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 long long read() {
  long long s = 0, t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch ^ 48), ch = getchar();
  return s * t;
}
const int N = 2e5 + 5;
const int P = 1e9 + 7;
int n, Ans;
long long T, s[N];
int p[N], fac[N], inv[N];
int Mod(int x) { return x >= P ? x - P : x; }
int Ksm(int x, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = 1ll * res * x % P;
    x = 1ll * x * x % P, k >>= 1;
  }
  return res;
}
void Pre(int n) {
  p[0] = 1;
  for (int i = 1; i <= n; i++) p[i] = 1ll * p[i - 1] * (P + 1) / 2 % P;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  inv[n] = Ksm(fac[n], P - 2);
  for (int i = n; i >= 1; i--) inv[i - 1] = 1ll * inv[i] * i % P;
}
int C(int n, int m) {
  return (n < m) ? 0 : 1ll * fac[n] * inv[m] % P * inv[n - m] % P;
}
int main() {
  n = read(), T = read(), Pre(n);
  for (int i = 1; i <= n; i++) s[i] = s[i - 1] + read();
  int x = 0, y = min(T, 1ll * n), res = 1;
  for (int i = 1; i <= n; i++)
    if (s[i] <= T) {
      while (x < i) res = Mod(2 * res % P + P - C(x, y)), x++;
      while (y > T - s[i]) res = Mod(res + P - C(x, y)), y--;
      Ans = Mod(Ans + 1ll * p[i] * res % P);
    }
  cout << Ans;
  return 0;
}
