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 long long N = 200069;
const long long mod = 1000000007;
const long long i2 = 500000004;
long long fac[N];
long long inv(long long x) {
  long long ans = 1;
  int p = mod - 2;
  while (p) {
    if (p & 1) ans = ans * x % mod;
    x = x * x % mod;
    p /= 2;
  }
  return ans;
}
long long binom(int n, int k) {
  return fac[n] * inv(fac[k] * fac[n - k] % mod) % mod;
}
int main() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = fac[i - 1] * i % mod;
  int n;
  long long T;
  cin >> n >> T;
  long long E = 0;
  long long p2 = 1;
  long long way = 1;
  int at = 0;
  for (int i = 1; i <= n; i++) {
    long long dT;
    cin >> dT;
    T -= dT + 1;
    way = way * 2 % mod;
    p2 = p2 * i2 % mod;
    if (++at != i) way = (way + binom(i - 1, at)) % mod;
    while (T < 0 && at >= 0) {
      way = (way - binom(i, at)) % mod;
      T++;
      at--;
    }
    E = (E + way * p2) % mod;
  }
  if (E < 0) E += mod;
  cout << E << '\n';
}
