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 long long MAXN = 2e5 + 10;
long long n, T, fac[MAXN], a[MAXN], inv[MAXN], b[MAXN], ans, mod = 1e9 + 7;
long long read() {
  long long sss = 0, fff = 1;
  char ccc;
  ccc = getchar();
  while (ccc < '0' || ccc > '9') {
    if (ccc == '-') fff = -1;
    ccc = getchar();
  }
  while (ccc >= '0' && ccc <= '9')
    sss = (sss << 1) + (sss << 3) + (ccc ^ '0'), ccc = getchar();
  return sss * fff;
}
long long ksm(long long base, long long k) {
  long long sum = 1;
  while (k) {
    if (k & 1) sum = sum * base % mod;
    base = base * base % mod;
    k >>= 1;
  }
  return sum;
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  return fac[n] * inv[n - m] % mod * inv[m] % mod;
}
int main() {
  n = read();
  T = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  b[n] = ksm(2, n);
  b[n] = ksm(b[n], mod - 2);
  for (long long i = n - 1; i >= 0; i--) b[i] = b[i + 1] * 2 % mod;
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  inv[n] = ksm(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  long long now = 0;
  bool flag = true;
  long long TT, sum = 0, CCC;
  while (now <= n && T >= a[now + 1]) {
    now++, T -= a[now];
    if (now <= T)
      ans = (ans + 1) % mod;
    else {
      if (flag) {
        TT = T;
        CCC = now;
        for (long long i = 0; i <= TT; i++) sum = (sum + C(now, i)) % mod;
      } else
        sum = (sum * 2 % mod - C(now - 1, TT) + mod) % mod;
      flag = false;
      while (TT > T) {
        sum = (sum - C(now, TT) + mod) % mod;
        TT--;
      }
      ans = (ans + sum * b[now] % mod) % mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
