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 = 200005;
const long long MOD = 1000000007;
int n;
long long T;
int t[MAXN];
long long done[MAXN];
long long mpow(long long bs, long long ep) {
  long long res = 1;
  while (ep) {
    if (ep & 1) {
      res = res * bs % MOD;
    }
    bs = bs * bs % MOD;
    ep >>= 1;
  }
  return res;
}
struct Combination {
  int n = -1, k = -1;
  long long res = 0;
  long long fac[MAXN], ifac[MAXN];
  long long cob(long long x, long long r) {
    assert(r <= x);
    return fac[x] * ifac[x - r] % MOD * ifac[r] % MOD;
  }
  void build() {
    fac[0] = ifac[0] = 1;
    for (int i = 1; i < MAXN; i++) {
      fac[i] = fac[i - 1] * i % MOD;
      ifac[i] = mpow(fac[i], MOD - 2);
    }
  }
  Combination() { build(); }
  long long count(int _n, int _k) {
    if (n == -1) {
      n = _n;
      k = _k;
      for (int i = 0; i <= _k; i++) {
        res = (res + cob(n, i)) % MOD;
      }
    } else {
      res = (res * 2 - cob(n, k)) % MOD;
      if (res < 0) {
        res += MOD;
      }
      n++;
      assert(n == _n);
      while (_k > k) {
        res += cob(n, k + 1);
        res %= MOD;
        k++;
      }
      while (_k < k) {
        res -= cob(n, k);
        res = (res + MOD) % MOD;
        k--;
      }
      assert(k == _k);
    }
    return res;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  cin >> n >> T;
  for (int i = 0; i < n; i++) {
    cin >> t[i];
  }
  long long pre = 0, base = 0, ans = 0;
  long long inv2 = mpow(2, MOD - 2);
  Combination cob;
  for (int i = 0; i < n; i++) {
    base = (base + t[i]);
    if (base > T) {
      break;
    }
    long long del = T - base;
    long long alive = 1;
    if (del < i + 1) {
      alive = cob.count(i + 1, del);
      ;
      alive = alive * mpow(inv2, i + 1) % MOD;
      ;
    }
    ans = (ans + alive) % MOD;
  }
  cout << ans << '\n';
}
