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 tt[200005];
long long t_sum2[200005];
long long rev_pow_2[200005];
long long jie[200005];
long long rev_jie[200005];
long long kpow(long long a, long long b) {
  long long ans = 1;
  a = a % mod;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b = b >> 1;
  }
  return ans;
}
void init() {
  long long rev_2 = kpow(2, mod - 2);
  rev_pow_2[1] = rev_2;
  jie[1] = 1;
  for (int i = 2; i < 200005; ++i) {
    rev_pow_2[i] = rev_pow_2[i - 1] * rev_2 % mod;
    jie[i] = jie[i - 1] * i % mod;
  }
  rev_jie[200005 - 1] = kpow(jie[200005 - 1], mod - 2);
  for (int i = 200005 - 2; i >= 1; --i)
    rev_jie[i] = rev_jie[i + 1] * (i + 1) % mod;
}
long long C(int a, int b) {
  if (a == b || b == 0) return 1;
  long long ans = jie[a] * rev_jie[a - b] % mod * rev_jie[b] % mod;
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  int n;
  long long T;
  cin >> n >> T;
  t_sum2[0] = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> tt[i];
    t_sum2[i] = t_sum2[i - 1] + tt[i] + 1;
  }
  int pos2 = upper_bound(t_sum2, t_sum2 + n + 1, T) - t_sum2 - 1;
  long long ans = 0;
  long long now_p = 1;
  for (int i = pos2;; ++i) {
    if (i == n) {
      ans += i * now_p % mod;
      ans %= mod;
      break;
    }
    long long cur_p = 0;
    int ii = 0;
    if (t_sum2[i] > T) ii = t_sum2[i] - T;
    for (; ii <= i && t_sum2[i] - ii + tt[i + 1] + 1 > T; ++ii) {
      if (t_sum2[i] - ii + tt[i + 1] <= T) {
        cur_p += C(i, ii) * rev_pow_2[i + 1] % mod;
        cur_p = cur_p % mod;
        break;
      }
      cur_p += C(i, ii) * rev_pow_2[i] % mod;
      cur_p = cur_p % mod;
    }
    ans += i * cur_p % mod;
    ans %= mod;
    now_p = (now_p - cur_p + mod) % mod;
    if (now_p == 0) break;
  }
  cout << ans << endl;
  return 0;
}
