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() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const int N = 3e5 + 5, mod = 1e9 + 7;
inline int inc(int x, int y) {
  x += y - mod;
  return x + (x >> 31 & mod);
}
inline int del(int x, int y) {
  x -= y;
  return x + (x >> 31 & mod);
}
inline int mul(int x, int y) { return 1ll * x * y % mod; }
inline void Inc(int &x, int y) { x = inc(x, y); }
inline void Del(int &x, int y) { x = del(x, y); }
inline void Mul(int &x, int y) { x = mul(x, y); }
inline long long Pow(long long x, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res = res * x % mod;
    x = x * x % mod;
    k >>= 1;
  }
  return res;
}
int inv[N], fac[N], inv2[N];
inline void init(int n) {
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = mul(fac[i - 1], i);
  inv[n] = Pow(fac[n], mod - 2);
  for (int i = n - 1; i >= 1; i--) inv[i] = mul(inv[i + 1], i + 1);
  int fac2 = 1;
  for (int i = 1; i <= n; i++) Inc(fac2, fac2);
  inv2[n] = Pow(fac2, mod - 2);
  for (int i = n - 1; i >= 1; i--) inv2[i] = inc(inv2[i + 1], inv2[i + 1]);
}
inline int C(int n, int m) {
  if (n < m) return 0;
  return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}
int n, ans;
long long s, T;
int main() {
  n = read();
  T = read();
  init(n);
  int j = n, sum = 1;
  for (int i = 1; i <= n; i++) {
    int x = read();
    s += x;
    if (s > T) continue;
    int l = min(1ll * n, T - s);
    sum = del(inc(sum, sum), C(i - 1, j));
    while (j > l) Del(sum, C(i, j)), j--;
    Inc(ans, mul(inv2[i], sum));
  }
  printf("%d\n", ans);
  return 0;
}
