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 readint() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int cys = 1000000007, inv2 = (cys + 1) / 2;
long long n, T, ans;
long long fac[200005], inv[200005], a[200005], s[200005];
long long mod(long long x) { return x >= cys ? x - cys : x; }
long long C(long long x, long long y) {
  return x < y ? 0 : fac[x] * inv[y] % cys * inv[x - y] % cys;
}
long long qpow(long long x, long long p) {
  long long ret = 1;
  for (; p; p >>= 1, x = x * x % cys)
    if (p & 1) ret = ret * x % cys;
  return ret;
}
int main() {
  n = readint();
  T = readint();
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % cys;
  inv[n] = qpow(fac[n], cys - 2);
  for (int i = n - 1; i >= 1; i--) inv[i] = inv[i + 1] * (i + 1) % cys;
  for (int i = 1; i <= n; i++) a[i] = readint(), s[i] = s[i - 1] + a[i];
  long long nw = n, nm = 0, res = 1;
  for (int i = 1; i <= n; i++) {
    long long x = min(n, T - s[i]);
    if (x < 0) break;
    while (nw > x) res = mod(res + cys - C(nm, nw)), nw--;
    while (nm < i) res = mod(mod(res << 1) + cys - C(nm, nw)), nm++;
    ans = mod(ans + res * qpow(inv2, i) % cys);
  }
  cout << ans << endl;
  return 0;
}
