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>
const int MOD = 1e9 + 7;
inline int addmod(int a, int b) { return (a + b) % MOD; }
inline int mulmod(long long a, int b) { return (a * b) % MOD; }
inline int powmod(int a, int x) {
  long long cur = a, res = 1;
  while (x) {
    if (x & 1) {
      res = (res * cur) % MOD;
    }
    x >>= 1;
    cur = (cur * cur) % MOD;
  }
  return res;
}
const int INV_2 = powmod(2, MOD - 2);
const int MAXN = 2e5 + 5;
int t[MAXN];
int fac[MAXN], inv_fac[MAXN];
int com(int n, int m) {
  return mulmod(fac[n], mulmod(inv_fac[m], inv_fac[n - m]));
}
int main() {
  fac[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    fac[i] = mulmod(fac[i - 1], i);
  }
  inv_fac[MAXN - 1] = powmod(fac[MAXN - 1], MOD - 2);
  for (int i = MAXN - 2; i >= 0; i--) {
    inv_fac[i] = mulmod(inv_fac[i + 1], i + 1);
  }
  int n;
  long long T;
  scanf("%d%lld", &n, &T);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &t[i]);
  }
  long long rem = T;
  int prefix = 1, pos = 0;
  int ans = 0;
  int p2 = 1;
  for (int i = 1; i <= n; i++) {
    rem -= t[i];
    if (rem < 0) break;
    while (pos < i - 1 && pos < rem - 1) {
      prefix = addmod(prefix, mulmod(p2, com(i - 1, ++pos)));
    }
    while (pos > rem - 1) {
      prefix = addmod(prefix, MOD - mulmod(p2, com(i - 1, pos--)));
    }
    ans = addmod(ans, prefix);
    if (rem <= i - 1) {
      ans = addmod(ans, mulmod(p2, mulmod(INV_2, com(i - 1, rem))));
    }
    prefix = addmod(prefix, MOD - mulmod(mulmod(p2, com(i - 1, pos)), INV_2));
    p2 = mulmod(p2, INV_2);
  }
  printf("%d\n", ans);
  return 0;
}
