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>
#pragma GCC optimize("-O3")
using namespace std;
const long long mod = 1e9 + 7;
struct factorial {
  int n;
  long long mod;
  int *a;
  int powm(int a, int b) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = int(1ll * ans * a % mod);
      a = int(1ll * a * a % mod);
      b >>= 1;
    }
    return ans;
  }
  int inv(int i) { return powm(i, int(mod) - 2); }
  factorial(int m, long long mod, int inverse) : n(m + 1), mod(mod) {
    if (inverse == 0) {
      a = new int[n];
      a[0] = 1;
      a[1] = 1;
      for (int i = 2; i < n; i++) {
        a[i] = int(1ll * i * a[i - 1] % mod);
      }
    } else {
      a = new int[n];
      a[0] = inv(1);
      a[1] = inv(1);
      for (int i = 2; i < n; i++) {
        a[i] = int(1ll * inv(i) * a[i - 1] % mod);
      }
    }
  }
};
const int N = 2e5 + 7;
int n, p1, p2;
long long tot;
long long sum;
long long tim, lsum;
int ans = 1;
int cr;
factorial fo(N, mod, 0), foinv(N, mod, 1);
int comb(int n, int r) {
  if (r < 0 || r > n) return 0;
  if (r == 0 || r == n) return 1;
  return 1ll * fo.a[n] * foinv.a[r] % mod * foinv.a[n - r] % mod;
}
void solve() {
  cin >> n >> tot;
  cin >> tim;
  sum = lsum = tim;
  ans = 0;
  for (int i = 1; i < n; i++) {
    cin >> tim;
    sum += tim;
    p1 = min(max(tot - sum + 1, 0ll), 1ll * i + 1);
    p2 = max(min(tot - lsum, 1ll * i), -1ll);
    cr = 0;
    for (int j = p1; j <= p2; j++) {
      cr += comb(i, j);
      cr %= mod;
    }
    ans += 1ll * fo.powm(foinv.a[2], i + 1) * cr % mod * i % mod;
    ans %= mod;
    p1 = min(max(tot - sum, 0ll), 1ll * i + 1);
    p2 = max(min(tot - lsum, 1ll * i), -1ll);
    cr = 0;
    for (int j = p1; j <= p2; j++) {
      cr += comb(i, j);
      cr %= mod;
    }
    ans += 1ll * fo.powm(foinv.a[2], i + 1) * cr % mod * i % mod;
    ans %= mod;
    lsum = sum;
  }
  sum = max(lsum + 1, tot + 1);
  p1 = min(max(tot - sum + 1, 0ll), 1ll * n + 1);
  p2 = max(min(tot - lsum, 1ll * n), -1ll);
  cr = 0;
  for (int j = p1; j <= p2; j++) {
    cr += comb(n, j);
    cr %= mod;
  }
  ans += 1ll * fo.powm(foinv.a[2], n) * cr % mod * n % mod;
  ans %= mod;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
