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 int maxn = 2 * 1e5 + 10;
const long long mod = 1e9 + 7;
long long n, T;
long long t[maxn];
long long fac[maxn];
long long invfac[maxn];
long long mul(long long a, long long b) { return a * b % mod; }
long long add(long long a, long long b) { return ((a + b) % mod + mod) % mod; }
long long fp(long long a, long long p) {
  if (p == 0) return 1;
  long long h = fp(a, p / 2);
  h = mul(h, h);
  if (p & 1) h = mul(h, a);
  return h;
}
long long C(long long n, long long k) {
  if (k > n || n < 0) return 0;
  return mul(mul(fac[n], invfac[k]), invfac[n - k]);
}
long long inv(long long a) { return fp(a, mod - 2); }
int main() {
  fac[0] = fac[1] = invfac[0] = invfac[1] = 1;
  for (int i = 2; i < maxn; i++) {
    fac[i] = mul(fac[i - 1], i);
    invfac[i] = mul(invfac[i - 1], inv(i));
  }
  long long inv2 = inv(2);
  cin >> n >> T;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
  }
  long long sum = 0, d = 1, psum = 1, ans = 0;
  int k = 0;
  for (int i = 1; i <= n; i++) {
    sum += t[i];
    if (sum > T) break;
    int ext = min((long long)i, T - sum);
    psum = mul(psum, 2);
    k++;
    psum = add(psum, C(i - 1, k));
    while (k > ext) {
      psum = add(psum, -C(i, k--));
    }
    d = mul(d, inv2);
    ans = add(ans, mul(psum, d));
  }
  cout << ans;
}
