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 n, m, i, j, fac[200005], inv[200005], fi[200005], a[200005],
    pw[200005], ans, mod = 1e9 + 7, lstx, lsty, tmp;
long long qp(long long x, long long y) {
  long long z = 1;
  while (y) {
    if (y & 1) {
      z = 1ll * z * x % mod;
    }
    x = 1ll * x * x % mod;
    y /= 2;
  }
  return z;
}
long long c(long long x, long long y) {
  return fac[x] * fi[y] % mod * fi[x - y] % mod;
}
long long solve(long long x, long long y) {
  while (lstx < x) {
    tmp = (tmp * 2 - c(lstx, lsty) + mod) % mod;
    lstx++;
  }
  while (lsty < y) {
    lsty++;
    tmp = (tmp + c(lstx, lsty) + mod) % mod;
  }
  while (lsty > y) {
    tmp = (tmp - c(lstx, lsty) + mod) % mod;
    lsty--;
  }
  return tmp;
}
int main() {
  fac[0] = fac[1] = inv[1] = fi[0] = fi[1] = 1;
  for ((i) = (2); (i) <= (200000); (i)++) {
    fac[i] = 1ll * fac[i - 1] * i % mod;
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    fi[i] = 1ll * fi[i - 1] * inv[i] % mod;
  }
  scanf("%lld%lld", &n, &m);
  pw[0] = 1;
  for ((i) = 1; (i) <= (n); (i)++) {
    pw[i] = 2ll * pw[i - 1] % mod;
    scanf("%lld", &a[i]);
    a[i] += a[i - 1];
  }
  lstx = 1;
  lsty = 0;
  tmp = 1;
  for ((i) = 1; (i) <= (n); (i)++) {
    long long t = min(i, m - a[i]);
    if (t < 0) continue;
    ans = (ans + pw[n - i] * solve(i, t)) % mod;
  }
  cout << 1ll * ans * qp(pw[n], mod - 2) % mod << endl;
  return 0;
}
