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 N = 2e5 + 1, mod = 1e9 + 7;
inline long long binpow(long long a, long long s) {
  long long ans = 1;
  while (s) {
    if (s & 1) ans = ans * a % mod;
    a = a * a % mod;
    s >>= 1;
  }
  return ans;
}
inline long long inv(long long a) { return binpow(a, mod - 2); }
int n;
long long t;
long long a[N];
long long sum[N];
long long ans;
long long fact[N], inv_fact[N], pow2[N], inv_pow2[N];
void init() {
  long long inv_2 = inv(2);
  fact[0] = 1;
  inv_fact[0] = 1;
  pow2[0] = 1;
  inv_pow2[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = fact[i - 1] * i % mod;
    inv_fact[i] = inv(fact[i]);
    pow2[i] = pow2[i - 1] * 2 % mod;
    inv_pow2[i] = inv_pow2[i - 1] * inv_2 % mod;
  }
}
long long C(long long n, long long k) {
  if (n < k) return 0ll;
  return fact[n] * inv_fact[k] % mod * inv_fact[n - k] % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  init();
  cin >> n >> t;
  for (int i = 0; i < n; ++i) cin >> a[i];
  if (a[0] > t) {
    cout << "0\n";
    return 0;
  }
  sum[0] = a[0];
  for (int i = 1; i < n; ++i) sum[i] = sum[i - 1] + a[i];
  long long x = (a[0] + 1 > t ? 0 : 1), p = 1 + x;
  ans = (x ? 1 : inv_pow2[1]);
  for (int i = 1; i < n; ++i) {
    p = (p * 2 + C(i, ++x)) % mod;
    while (x >= 0 && sum[i] + x > t) p = (p + mod - C(i + 1, x--)) % mod;
    ans = (ans + p * inv_pow2[i + 1]) % mod;
  }
  cout << ans << '\n';
}
