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>
#pragma GCC optimize("O3")
using namespace std;
const int MAX = 3e5 + 5;
const long long MAX2 = 11;
const long long MOD = 1000000007;
const long long MOD2 = 1000000006;
const long long INF = 2e18;
const int dr[] = {1, 0, -1, 0, 1, 1, -1, -1, 0};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1, 0};
const double pi = acos(-1);
const double EPS = 1e-9;
const int block = 316;
long long n, x[MAX], t, f[MAX], pre[MAX], ans, tmp, inv, dv;
inline long long pw(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = ret * x % MOD;
    x = x * x % MOD;
    y >>= 1;
  }
  return ret;
}
inline void que(long long a, long long b, long long k) {
  a = max(0ll, a), b = min(k, b);
  if (a > b) return;
  for (long long i = a; i <= b; ++i)
    tmp = (tmp + f[k] * pre[i] % MOD * pre[k - i]) % MOD;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> t;
  f[0] = pre[0] = 1;
  for (long long i = 1; i <= n; ++i)
    cin >> x[i], x[i] += x[i - 1], f[i] = f[i - 1] * i % MOD,
                                   pre[i] = pw(f[i], MOD - 2);
  inv = pw(2, MOD - 2), dv = inv, x[n + 1] = INF;
  for (long long i = 1; i <= n; ++i) {
    tmp = 0;
    if (t - x[i + 1] >= 0 && t - x[i + 1] <= i)
      tmp = f[i] * pre[t - x[i + 1]] % MOD * pre[i - t + x[i + 1]] % MOD * inv %
            MOD;
    que(t - x[i + 1] + 1, t - x[i], i);
    ans = (ans + i * tmp % MOD * dv) % MOD;
    dv = dv * inv % MOD;
  }
  cout << ans << '\n';
  return 0;
}
