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("Ofast", "unroll-loops")
using namespace std;
const int maxn = 2e5 + 5;
const int iinf = 1 << 29;
const long long inf = 1ll << 60;
const long long mod = 1e9 + 7;
void GG() {
  cout << "No\n";
  exit(0);
}
long long mpow(long long a, long long n) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) re = re * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return re;
}
long long inv(long long b) {
  if (b == 1) return b;
  return (mod - mod / b) * inv(mod % b) % mod;
}
long long fac[maxn];
void build() {
  fac[0] = 1;
  for (int i = (1); i < (maxn); i++) {
    fac[i] = fac[i - 1] * i % mod;
  }
}
long long C(long long a, long long b) {
  return fac[a] * inv(fac[a - b]) % mod * inv(fac[b]) % mod;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  build();
  int n;
  long long t;
  cin >> n >> t;
  vector<long long> a(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  long long nx = 0;
  long long p2 = 1;
  long long x = 0;
  long long re = 0;
  long long sum = 0;
  for (int i = 0; i < (n); i++) {
    p2 = p2 * 2 % mod;
    sum += a[i];
    if (sum > t) break;
    if (sum + i + 1 > t) {
      if (nx == 0)
        nx = p2, x = i + 2;
      else {
        nx = (mod + nx * 2 - C(i, x - 1)) % mod;
      }
      while (sum + x > 1 + t) {
        nx = (mod + nx - C(i + 1, x - 1)) % mod;
        x--;
      }
      re += nx * inv(p2) % mod;
      re %= mod;
    } else {
      re++;
      re %= mod;
    }
  }
  cout << re << '\n';
}
