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 int MAXN = 200005;
const long long MOD = 1000000007;
long long N, T, ts[MAXN], pref[MAXN];
long long fac[MAXN], inv[MAXN];
long long C(long long n, long long k) {
  return fac[n] * inv[k] % MOD * inv[n - k] % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  fac[0] = inv[0] = fac[1] = inv[1] = 1;
  for (int i = 2; i < MAXN; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = -MOD / i * inv[MOD % i] % MOD;
  }
  for (int i = 2; i < MAXN; i++) {
    inv[i] = inv[i - 1] * inv[i] % MOD;
  }
  int k = 1;
  long long ans = 0, L = 2, deno = inv[2];
  cin >> N >> T;
  for (int i = 1; i <= N; i++) {
    cin >> ts[i];
    pref[i] = ts[i] + pref[i - 1];
    if (pref[i] > T) {
      break;
    }
    long long most = min((long long)i, T - pref[i]);
    while (k < most) {
      k++;
      L = (L + C(i, k)) % MOD;
    }
    while (k > most) {
      L = (L - C(i, k)) % MOD;
      k--;
    }
    ans = (ans + deno * L) % MOD;
    deno = deno * inv[2] % MOD;
    L = (L * 2 - C(i, k)) % MOD;
  }
  cout << (ans + MOD) % MOD << '\n';
  return 0;
}
