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 c = getchar();
  long long x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
const long long Mod = 1e9 + 7;
const long long M = 2e5 + 10;
long long n, t, a[M], s[M], ans, f[M], p[M];
long long fac[M], ifac[M];
long long poww(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % Mod;
    a = a * a % Mod, b >>= 1;
  }
  return res;
}
long long inv(long long x) { return poww(x, Mod - 2); }
void init(long long x) {
  fac[0] = ifac[0] = 1;
  for (long long i = 1; i <= x; i++)
    fac[i] = fac[i - 1] * i % Mod, ifac[i] = inv(fac[i]);
}
long long C(long long m, long long n) {
  if (n > m) return 0;
  return fac[m] * ifac[n] % Mod * ifac[m - n] % Mod;
}
signed main() {
  n = read(), t = read();
  init(2e5);
  for (long long i = 1; i <= n; i++) a[i] = read(), s[i] = s[i - 1] + a[i];
  for (long long i = 1; i <= n; i++) {
    long long x = min(i, t - s[i]);
    p[i] = t - s[i];
    if (x < 0) break;
    if (i == 1) {
      for (long long j = 0; j <= x; j++) f[i] = (f[i] + C(i, j)) % Mod;
    } else {
      f[i] = (2 * f[i - 1] - C(i - 1, p[i - 1])) % Mod;
      for (long long j = min(p[i - 1], i); j > p[i]; j--)
        f[i] = (f[i] - C(i, j) + Mod) % Mod;
    }
  }
  for (long long i = n; i >= 1; i--) ans = (ans + f[i] * poww(2, n - i)) % Mod;
  ans = ans * inv(poww(2, n)) % Mod;
  cout << ans << endl;
  return 0;
}
