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 = 2e5 + 14, mod = 1e9 + 7;
int n;
long long T, t[maxn];
int po2[maxn], fac[maxn], rfac[maxn];
int rev(int a) {
  int ans = 1;
  for (int b = mod - 2; b; b >>= 1, a = (long long)a * a % mod)
    if (b & 1) ans = (long long)ans * a % mod;
  return ans;
}
void prep() {
  fac[0] = rfac[0] = po2[0] = 1;
  for (int i = 1; i <= n; i++) {
    po2[i] = po2[i - 1] * 2 % mod;
    fac[i] = (long long)i * fac[i - 1] % mod;
    rfac[i] = rev(fac[i]);
  }
}
int c(int n, int r) {
  return (long long)fac[n] * rfac[r] % mod * rfac[n - r] % mod;
}
int sumc(int n, long long r) {
  if (r < 0) return 0;
  if (r >= n) return po2[n];
  static int curn = n, curr = 0, curans = 0;
  assert(curn >= n && curr <= r + 1);
  while (curn > n) {
    curn--;
    curans = (curans + c(curn, curr - 1)) % mod;
    curans = (long long)curans * rev(2) % mod;
  }
  while (curr <= r) curans = (curans + c(n, curr++)) % mod;
  return curans;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> T;
  prep();
  for (int i = 0; i < n; i++) {
    cin >> t[i + 1];
    t[i + 1] += t[i];
  }
  int ans = 0, last = 0;
  for (int i = n; i > 0; i--) {
    int prob =
        (long long)sumc(i, min<long long>(T - t[i], i)) * rev(po2[i]) % mod;
    ans = (ans + (long long)(prob - last + mod) * i) % mod;
    last = prob;
  }
  cout << ans << '\n';
}
