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;
const int MAXN = 2e5 + 5;
long long fac[MAXN], ifac[MAXN];
long long mpow(long long b, long long e) {
  long long res = 1;
  for (long long k = 1; k <= e; k <<= 1) {
    if (e & k) res = (res * b) % MOD;
    b = (b * b) % MOD;
  }
  return res;
}
long long inv(long long b) { return mpow(b, MOD - 2); }
long long binom(long long n, long long k) {
  long long res = fac[n];
  res = (res * ifac[k]) % MOD;
  res = (res * ifac[n - k]) % MOD;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  fac[0] = 1;
  for (long long i = 1; i < MAXN; i++) fac[i] = (i * fac[i - 1]) % MOD;
  ifac[MAXN - 1] = inv(fac[MAXN - 1]);
  for (long long i = MAXN - 1; i > 0; i--) ifac[i - 1] = (i * ifac[i]) % MOD;
  int n;
  long long T;
  cin >> n >> T;
  vector<long long> v(n), pf(n + 1);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    pf[i + 1] = pf[i] + v[i];
  }
  long long ans = 0;
  for (int m = 0; m <= n; m++) {
    if (pf[m] > T) break;
    if (m < n && pf[m + 1] + m + 1 <= T) continue;
    long long k = 0;
    if (m < n) k = max(0LL, T - pf[m + 1]);
    for (; k <= m && pf[m] + k <= T; k++) {
      long long tmp = (binom(m, k) * inv(mpow(2, m))) % MOD;
      tmp = (tmp * m) % MOD;
      if (m < n && pf[m + 1] + k <= T) {
        tmp = (tmp * inv(2)) % MOD;
      }
      ans = (ans + tmp) % MOD;
    }
  }
  cout << ans << '\n';
}
