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;
long long a[200010], s[200010];
const long long mod = 1000000007, hf = mod + 1 >> 1;
long long _2[200010];
long long fac[200010], inv[200010], facinv[200010];
long long C(int a, int b) {
  return fac[a] * facinv[b] % mod * facinv[a - b] % mod;
}
int main() {
  _2[0] = 1;
  int n;
  long long T;
  scanf("%d%lld", &n, &T);
  int i;
  fac[0] = facinv[0] = 1;
  fac[1] = facinv[1] = inv[1] = 1;
  for (i = 2; i <= n; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
    facinv[i] = facinv[i - 1] * inv[i] % mod;
  }
  for (i = 1; i <= n; i++) {
    _2[i] = _2[i - 1] * hf % mod;
    scanf("%d", &a[i]);
    s[i] = s[i - 1] + a[i];
  }
  long long L, R;
  long long ans = 0;
  for (i = 1; i < n; i++) {
    long long sum = 0;
    L = T - a[i + 1] + 1;
    R = T;
    L -= s[i];
    R -= s[i];
    if (L > i || R < 0) goto br;
    L = max(L, 0ll);
    R = min(R, (long long)i);
    for (int j = L; j <= R; j++) {
      sum += C(i, j);
    }
    sum %= mod;
    ans += sum * _2[i + 1] % mod * i % mod;
  br:;
    a[i + 1]++;
    L = T - a[i + 1] + 1;
    R = T;
    L -= s[i];
    R -= s[i];
    if (L > i || R < 0) continue;
    L = max(L, 0ll);
    R = min(R, (long long)i);
    sum = 0;
    for (int j = L; j <= R; j++) {
      sum += C(i, j);
    }
    sum %= mod;
    ans += sum * _2[i + 1] % mod * i % mod;
    a[i + 1]--;
    ans %= mod;
  }
  long long tmp = T - s[n];
  if (tmp < 0) {
    printf("%lld\n", ans);
    return 0;
  }
  tmp = min(tmp, (long long)n);
  long long sum = 0;
  for (i = 0; i <= tmp; i++) {
    sum += C(n, i);
  }
  sum %= mod;
  ans += sum * _2[n] % mod * n % mod;
  ans %= mod;
  printf("%lld\n", ans);
  return 0;
}
