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 mod = (long long)1e9 + 7;
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return (binpow(a, n - 1) * a) % mod;
  else {
    long long b = binpow(a, n / 2);
    return (b * b) % mod;
  }
}
long long add(long long x, long long y) { return (x + y) % mod; }
long long sub(long long x, long long y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
long long mult(long long x, long long y) { return (x * y) % mod; }
long long rev(long long x) { return binpow(x, mod - 2); }
int n;
long long T;
long long t[400400], second[400400], fact[400400], rev_fact[400400], m[400400],
    pw2[400400], rev_pw2[400400];
long long cnk(long long n, long long k) {
  long long result = mult(fact[n], mult(rev_fact[n - k], rev_fact[k]));
  return result;
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> T;
  fact[0] = 1;
  rev_fact[0] = 1;
  pw2[0] = 1;
  rev_pw2[0] = 1;
  for (int i = 1; i <= n; ++i) {
    cin >> t[i];
    second[i] = second[i - 1] + t[i];
    fact[i] = mult(i, fact[i - 1]);
    rev_fact[i] = rev(fact[i]);
    pw2[i] = mult(2, pw2[i - 1]);
    rev_pw2[i] = rev(pw2[i]);
  }
  long long sum = 0, ans = 0;
  long long rev2 = rev(2);
  for (int i = 1; i <= n; ++i) {
    m[i] = min((long long)i, T - second[i]);
    if (m[i] < 0) break;
    if (i > 1) {
      sum = mult(sum, 2);
      sum = add(sum, cnk(i, m[i - 1] + 1));
      sum = sub(sum, cnk(i - 1, 0));
      sum = add(sum, cnk(i, 0));
      sum = sub(sum, cnk(i - 1, m[i - 1]));
    } else {
      sum = cnk(1, 0) + cnk(1, 1);
    }
    long long ind = m[i - 1] + 1;
    while (ind > m[i]) {
      sum = sub(sum, cnk(i, ind));
      --ind;
    }
    ans = add(ans, mult(sum, rev_pw2[i]));
  }
  cout << ans << "\n";
  return 0;
}
