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 ON = 200000;
const int MAXN = 200005;
const long long ONE = 1;
const int MOD = 1000000007;
int N, t[MAXN];
long long F[MAXN];
long long T, Sum[MAXN], Ans;
long long f[MAXN] = {1}, fe[MAXN] = {1};
long long O[MAXN] = {1}, Oe[MAXN] = {1};
long long Sum_N, Sum_K;
long long quick_Pow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  if (y % 2) return (x * quick_Pow((x * x) % MOD, y / 2)) % MOD;
  return quick_Pow((x * x) % MOD, y / 2);
}
void Prepare() {
  for (int i = 1; i <= ON; i++) {
    f[i] = (f[i - 1] * i) % MOD;
    fe[i] = quick_Pow(f[i], MOD - 2);
    O[i] = (O[i - 1] * 2) % MOD;
    Oe[i] = quick_Pow(O[i], MOD - 2);
  }
}
long long C(long long x, long long y) {
  if (y > x) return 0;
  return (f[x] * ((fe[y] * fe[x - y]) % MOD)) % MOD;
}
long long work(long long n, long long k) {
  if (Sum_K == 0) {
    for (int i = 0; i <= k; i++) Sum_N = (Sum_N + C(n, i)) % MOD;
  } else {
    Sum_N = (Sum_N * 2 - C(n - 1, Sum_K) + MOD) % MOD;
    for (int i = Sum_K; i > k; i--) Sum_N = (Sum_N - C(n, i) + MOD) % MOD;
  }
  Sum_K = k;
  return Sum_N;
}
int main() {
  Prepare();
  scanf("%d%lld", &N, &T);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &t[i]);
    Sum[i] = Sum[i - 1] + t[i];
  }
  for (int i = 1; i <= N; i++) {
    if (Sum[i] > T) break;
    if (Sum[i] + i <= T) {
      F[i] = 1;
      continue;
    }
    F[i] = (work(i, T - Sum[i]) * Oe[i]) % MOD;
  }
  for (int i = 1; i <= N; i++) Ans = (Ans + F[i]) % MOD;
  printf("%lld\n", Ans);
}
