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 + 10, MOD = 1e9 + 7;
long long n, T, ni[MAXN], jie[MAXN], jie_ni[MAXN], A[MAXN], ans, sum;
long long sum_N, sum_K;
void Init() {
  ni[1] = 1;
  for (long long i = 2; i <= MAXN - 10; i++) {
    ni[i] = ((MOD - (MOD / i)) * ni[MOD % i] % MOD) % MOD;
  }
  jie_ni[0] = 1;
  for (long long i = 1; i <= MAXN - 10; i++) {
    jie_ni[i] = jie_ni[i - 1] * ni[i] % MOD;
  }
  jie[0] = 1;
  for (long long i = 1; i <= MAXN - 10; i++) {
    jie[i] = jie[i - 1] * i % MOD;
  }
}
long long Quick(long long x, long long y) {
  long long res = 1, a = x;
  while (y) {
    if (y & 1) res *= a, res %= MOD;
    a = a * a, a %= MOD;
    y /= 2;
  }
  return res;
}
long long C(long long x, long long y) {
  if (y == 0) return 1;
  if (x == y) return 1;
  return (jie[x] * jie_ni[y]) % MOD * jie_ni[x - y] % MOD;
}
long long work(long long N, long long K) {
  if (sum_N == 0) {
    sum_N = 1;
    for (long long i = 1; i <= K; i++) sum_N += C(N, i), sum_N %= MOD;
    sum_K = K;
    return sum_N;
  } else {
    sum_N = (sum_N * 2 - C(N - 1, sum_K) + MOD) % MOD;
    for (long long i = sum_K; i >= K + 1; i--) {
      sum_N -= C(N, i), sum_N = (sum_N + MOD) % MOD;
    }
    sum_K = K;
    return sum_N;
  }
}
signed main() {
  cin >> n >> T;
  Init();
  for (long long i = 1; i <= n; i++) cin >> A[i];
  long long BASE = 1;
  for (long long i = 1; i <= n; i++) {
    sum += A[i];
    BASE = BASE * 2;
    BASE %= MOD;
    if (T - sum >= i) {
      ans++;
      ans %= MOD;
    } else if (T - sum < 0) {
      break;
    } else {
      ans += (work(i, T - sum) * Quick(BASE, MOD - 2)) % MOD;
      ans %= MOD;
    }
  }
  cout << ans;
  return 0;
}
