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;
inline long long read() {
  char ch = getchar();
  long long nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  long long ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
long long fac[200005];
long long inv[200005];
long long a[200005];
long long C(long long x, long long y) {
  return fac[x] * inv[y] % 1000000007 * inv[x - y] % 1000000007;
}
long long qpow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return ans;
}
signed main() {
  long long n, T;
  cin >> n >> T;
  for (long long i = 1; i <= n; i++) {
    a[i] = read();
    a[i] += a[i - 1];
  }
  fac[0] = fac[1] = 1;
  inv[0] = inv[1] = 1;
  for (long long i = 2; i <= n; i++) {
    fac[i] = fac[i - 1] * i % 1000000007;
    inv[i] = (1000000007 - 1000000007 / i) * inv[1000000007 % i] % 1000000007;
  }
  for (long long i = 2; i <= n; i++) inv[i] = inv[i - 1] * inv[i] % 1000000007;
  long long ans = 0, tmp = 0;
  bool DO = true;
  long long siz = 0;
  for (long long i = 1; i <= n; i++) {
    if (T - a[i] >= i)
      ans = (ans + 1) % 1000000007;
    else {
      if (T - a[i] < 0) break;
      if (DO) {
        DO = false;
        siz = T - a[i];
        for (long long j = 0; j <= siz; j++) {
          tmp = (tmp + C(i, j)) % 1000000007;
        }
        ans = (ans + tmp * qpow(qpow(2, i), 1000000007 - 2) % 1000000007) %
              1000000007;
      } else {
        long long x = T - a[i];
        for (long long j = x + 1; j <= siz; j++) {
          tmp = (tmp - C(i - 1, j) + 1000000007) % 1000000007;
        }
        siz = x;
        tmp = tmp * 2 % 1000000007;
        tmp = (tmp - C(i - 1, x) + 1000000007) % 1000000007;
        ans = (ans + tmp * qpow(qpow(2, i), 1000000007 - 2) % 1000000007) %
              1000000007;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
