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 = 1000000007;
const long long MAX_N = 200001;
long long get_power(long long n, long long pow) {
  if (pow == 0) {
    return 1;
  }
  if (pow % 2) {
    return (get_power(n, pow - 1) * n) % MOD;
  }
  long long tmp = get_power(n, pow / 2);
  return (tmp * tmp) % MOD;
}
long long get_inverse(long long n) { return get_power(n, MOD - 2); }
long long fact[MAX_N], inv_fact[MAX_N];
void prepare(long long n) {
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % MOD;
  }
  for (long long i = 0; i <= n; i++) {
    inv_fact[i] = get_inverse(fact[i]);
  }
}
long long get_binom(long long n, long long k) {
  if (n < k || n < 0 || k < 0) {
    return 0;
  }
  return ((fact[n]) * ((inv_fact[k] * inv_fact[n - k]) % MOD)) % MOD;
}
long long solve() {
  long long n;
  long long T;
  cin >> n >> T;
  prepare(n);
  vector<long long> t(n);
  for (long long i = 0; i < n; i++) {
    cin >> t[i];
  }
  vector<long long> undivided_p(n + 1);
  long long pref_t = t[0], binom_sum = 0;
  long long m = max(min(T - pref_t, (long long)1), (long long)-1);
  for (long long k = 0; k <= m; k++) {
    binom_sum = (binom_sum + get_binom(1, k)) % MOD;
  }
  undivided_p[1] = binom_sum;
  for (long long i = 2; i <= n; i++) {
    pref_t += t[i - 1];
    long long new_m = max(min(T - pref_t, i), (long long)-1);
    for (long long k = m + 1; k <= new_m - 1; k++) {
      binom_sum = (binom_sum + get_binom(i - 1, k)) % MOD;
    }
    for (long long k = m; k > new_m - 1; k--) {
      binom_sum = (binom_sum + MOD - get_binom(i - 1, k)) % MOD;
    }
    binom_sum = (2 * binom_sum + get_binom(i - 1, new_m)) % MOD;
    undivided_p[i] = binom_sum;
    m = new_m;
  }
  long long result = 0;
  long long base = 2;
  for (long long i = 1; i <= n; i++) {
    result = (result + undivided_p[i] * get_inverse(base)) % MOD;
    base = (2 * base) % MOD;
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << solve() << '\n';
  return 0;
}
