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 maxn = 2e5 + 3;
const long long mod = 1e9 + 7;
long long t[maxn], fac[maxn], rev[maxn], pw[maxn];
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long ret = power(a, b / 2);
  ret = 1ll * ret * ret % mod;
  if (b & 1) ret = 1ll * ret * a % mod;
  return ret;
}
long long Choose(long long n, long long r) {
  return 1ll * fac[n] * rev[r] % mod * rev[n - r] % mod;
}
long long get(long long a, long long b) {
  return 1ll * a * power(b, mod - 2) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n;
  long long T;
  cin >> n >> T;
  pw[0] = 1;
  for (long long i = 1; i <= n; i++) pw[i] = pw[i - 1] * 2 % mod;
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  rev[n] = power(fac[n], mod - 2);
  for (long long i = n - 1; i >= 0; i--)
    rev[i] = 1ll * rev[i + 1] * (i + 1) % mod;
  for (long long i = 1; i <= n; i++) cin >> t[i];
  long long idx = 0;
  for (long long i = 1; i <= n; i++) {
    T -= t[i];
    if (T < i) {
      T += t[i];
      idx = i;
      break;
    }
    idx = i;
    if (i == n) return cout << n << endl, 0;
  }
  long long answer = idx - 1;
  T -= t[idx];
  long long bin = 0;
  for (long long i = 0; i <= T; i++) bin = (bin + Choose(idx, i)) % mod;
  answer = (answer + get(bin, pw[idx])) % mod;
  for (long long i = idx + 1; i <= n; i++) {
    if (T - t[i] < 0) break;
    bin = (2ll * bin - Choose(i - 1, T) + mod) % mod;
    for (long long j = T; j > T - t[i]; j--)
      bin = (bin - Choose(i, j) + mod) % mod;
    answer = (answer + get(bin, pw[i])) % mod;
    T -= t[i];
  }
  cout << answer << endl;
}
