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 mod = 1e9 + 7;
long long p(long long a, long long b) {
  if (b == 0) return 1;
  long long tmp = p(a, b / 2);
  if (b % 2 == 0) return (tmp * tmp) % mod;
  return ((tmp * tmp) % mod * a) % mod;
}
long long inv(long long x) { return p(x, mod - 2); }
long long n, T;
long long t[200001];
long long m[200001];
long long fact[200001];
long long re_fact[200001];
long long res_E = 0;
long long comb(long long n, long long k) {
  if (k > n) return 0;
  long long c = ((fact[n] * re_fact[k]) % mod * re_fact[n - k]) % mod;
  return c;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fact[0] = 1;
  for (long long i = 1; i <= 200000; i++) fact[i] = (fact[i - 1] * i) % mod;
  re_fact[200000] = inv(fact[200000]);
  for (long long i = 200000; i >= 1; i--)
    re_fact[i - 1] = (re_fact[i] * i) % mod;
  cin >> n >> T;
  long long sum_t = 0;
  long long pre_p = 1;
  for (long long i = 1; i <= n; i++) {
    cin >> t[i];
    sum_t += t[i];
    m[i] = min(T - sum_t, i);
    if (m[i] < 0) break;
    pre_p = ((2 * pre_p) % mod + comb(i - 1, m[i - 1] + 1)) % mod;
    if (m[i] != m[i - 1] + 1)
      for (long long j = m[i - 1] + 1; j > m[i]; j--)
        pre_p = ((pre_p - comb(i, j)) % mod + mod) % mod;
    res_E = (res_E + (inv(p(2, i)) * pre_p) % mod) % mod;
  }
  cout << res_E;
}
