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 mod = 1e9 + 7;
const int N = 2e5 + 1;
long long tp[N], p[N], t[N];
long long bp(long long x, int st) {
  if (st == 0)
    return 1;
  else if (st % 2 == 0)
    return bp((x * x) % mod, st / 2);
  else
    return (bp((x * x) % mod, st / 2) * x) % mod;
}
long long c(long long n, long long k) {
  return (((p[n] * bp(p[k], mod - 2)) % mod) * bp(p[n - k], mod - 2)) % mod;
}
int main() {
  long long i, j, k, n, T;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> T;
  long long ans = 0, sum = 0, r = 0, sumr = 1;
  tp[0] = 1;
  p[0] = 1;
  for (i = 1; i <= n; ++i) {
    tp[i] = (tp[i - 1] * 2) % mod;
    p[i] = (p[i - 1] * i) % mod;
    cin >> t[i];
    if (r >= 0) sumr = (2 * sumr - c(i - 1, r)) % mod;
    if (r + 1 <= T - sum) {
      sumr = (sumr + 1) % mod;
      ++r;
    }
    sum += t[i];
    while (r >= 0 && r > T - sum) {
      sumr = (sumr - c(i, r)) % mod;
      --r;
    }
    if (sum + i <= T)
      ans = (ans + 1) % mod;
    else if (sum <= T) {
      ans = (ans + sumr * bp(tp[i], mod - 2)) % mod;
    }
  }
  cout << (ans + mod) % mod;
}
