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 M = 200005;
const int MOD = 1e9 + 7;
long long read() {
  long long num = 0, flag = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  while (c >= '0' && c <= '9')
    num = (num << 3) + (num << 1) + (c ^ 48), c = getchar();
  return num * flag;
}
long long n, m, s, p, ans, a[M], fac[M], inv[M], pw[M];
void init(long long n) {
  fac[0] = fac[1] = inv[0] = inv[1] = pw[0] = 1;
  for (long long i = 2; i <= n; i++)
    inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
  for (long long i = 2; i <= n; i++) inv[i] = inv[i] * inv[i - 1] % MOD;
  for (long long i = 2; i <= n; i++) fac[i] = fac[i - 1] * i % MOD;
  for (long long i = 1; i <= n; i++) pw[i] = pw[i - 1] * 500000004 % MOD;
}
long long C(long long n, long long m) {
  if (n < m) return 0;
  return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}
signed main() {
  init(200000);
  n = read();
  m = read();
  for (long long i = 1; i <= n; i++) a[i] = a[i - 1] + read();
  s = 1;
  p = min(m, n);
  for (long long i = 1; i <= n; i++) {
    if (m - a[i] < 0) break;
    s = (2 * s - C(i - 1, p)) % MOD;
    long long np = min(m - a[i], n);
    for (long long j = p; j > np; j--) s = (s - C(i, j)) % MOD;
    p = np;
    ans = (ans + pw[i] * s) % MOD;
  }
  printf("%lld\n", (ans + MOD) % MOD);
}
