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>
template <typename T, typename _T>
inline bool chk_min(T &x, const _T y) {
  return y < x ? x = y, 1 : 0;
}
template <typename T, typename _T>
inline bool chk_max(T &x, const _T y) {
  return x < y ? x = y, 1 : 0;
}
const int P = 1e9 + 7;
const int N = 2e5 + 5;
int fac[N], ifac[N];
long long s[N];
int n;
long long T;
int Pow(int a, int p) {
  int res = 1;
  for (; p > 0; p >>= 1, a = 1ll * a * a % P)
    if (p & 1) res = 1ll * res * a % P;
  return res;
}
int inv(int a) { return Pow(a, P - 2); }
int pls(int x, int y) {
  x += y;
  if (x >= P) x -= P;
  return x;
}
int nCr(int n, int m) { return 1ll * fac[n] * ifac[m] % P * ifac[n - m] % P; }
int calc(int x, long long yl, long long yr) {
  chk_max(yl, 0);
  chk_min(yr, x);
  if (yl > yr) return 0;
  int res = 0;
  for (int i = (yl), iEND = (yr); i <= iEND; ++i) res = pls(res, nCr(x, i));
  return res;
}
int main() {
  fac[0] = ifac[0] = ifac[1] = 1;
  for (int i = (1), iEND = (N - 2); i <= iEND; ++i)
    fac[i] = 1ll * fac[i - 1] * i % P;
  for (int i = (2), iEND = (N - 2); i <= iEND; ++i)
    ifac[i] = 1ll * (P - P / i) * ifac[P % i] % P;
  for (int i = (2), iEND = (N - 2); i <= iEND; ++i)
    ifac[i] = 1ll * ifac[i - 1] * ifac[i] % P;
  scanf("%d%lld", &n, &T);
  for (int i = (1), iEND = (n); i <= iEND; ++i)
    scanf("%lld", &s[i]), s[i] += s[i - 1];
  int ans = 0;
  for (int i = (1), iEND = (n - 1); i <= iEND; ++i) {
    ans = pls(ans, 1ll * i * calc(i, T - s[i + 1] + 1, T - s[i]) % P *
                       Pow(2, n - i - 1) % P);
    ans = pls(ans, 1ll * i * calc(i, T - s[i + 1], T - s[i]) % P *
                       Pow(2, n - i - 1) % P);
  }
  ans = pls(ans, 1ll * n * calc(n, 0, T - s[n]) % P);
  printf("%lld\n", 1ll * ans * inv(Pow(2, n)) % P);
  return 0;
}
