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 + 5, Mod = 1e9 + 7;
long long fac[maxn], inv[maxn], pw[maxn];
long long power(long long b, long long e) {
  if (!e) return 1;
  long long tmp = power(b, e >> 1);
  tmp = 1ll * tmp * tmp % Mod;
  if (e & 1) tmp = 1ll * tmp * b % Mod;
  return tmp;
}
long long n, t[maxn], st[maxn], T, ans;
signed main() {
  fac[0] = inv[0] = 1;
  for (long long i = 1; i <= 2e5; i++)
    fac[i] = 1ll * fac[i - 1] * i % Mod, inv[i] = power(fac[i], Mod - 2);
  pw[0] = 1;
  for (long long i = 1; i <= 2e5; i++) pw[i] = 2ll * pw[i - 1] % Mod;
  cin >> n >> T;
  for (long long i = 1; i <= n; i++) cin >> t[i], st[i] = st[i - 1] + t[i];
  long long i;
  for (i = 1; i <= n && st[i] + i <= T; i++) ans++;
  long long c = -1;
  for (; i <= n && st[i] <= T; i++) {
    if (c == -1) {
      c = 0;
      for (long long j = 0; j <= T - st[i]; j++)
        c = (c + 1ll * fac[i] * inv[j] % Mod * inv[i - j] % Mod) % Mod;
    } else {
      c = (2ll * c -
           1ll * fac[i - 1] * inv[T - st[i - 1]] % Mod *
               inv[i - 1 - T + st[i - 1]] % Mod +
           Mod) %
          Mod;
      for (long long j = T - st[i - 1]; j > T - st[i]; j--)
        c = (c - 1ll * fac[i] * inv[j] % Mod * inv[i - j] % Mod + Mod) % Mod;
    }
    ans = (ans + 1ll * power(pw[i], Mod - 2) * c % Mod) % Mod;
  }
  cout << ans << endl;
  return 0;
}
