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>
const int Mod = 1000000007;
const int Inv2 = (Mod + 1) / 2;
const int MN = 200005;
inline int qPow(int b, int e) {
  int a = 1;
  for (; e; e >>= 1, b = (long long)b * b % Mod)
    if (e & 1) a = (long long)a * b % Mod;
  return a;
}
int Inv[MN], Fac[MN], iFac[MN];
inline int Binom(int N, int M) {
  if (M < 0 || M > N) return 0;
  return (long long)Fac[N] * iFac[M] % Mod * iFac[N - M] % Mod;
}
inline void Init(int N) {
  Inv[1] = 1;
  for (int i = 2; i <= N; ++i)
    Inv[i] = (long long)(Mod - Mod / i) * Inv[Mod % i] % Mod;
  Fac[0] = iFac[0] = 1;
  for (int i = 1; i <= N; ++i) {
    Fac[i] = (long long)Fac[i - 1] * i % Mod;
    iFac[i] = (long long)iFac[i - 1] * Inv[i] % Mod;
  }
}
int N, x, Ans, Now, X, P2;
long long T, S;
int main() {
  scanf("%d%lld", &N, &T);
  Init(N);
  Now = 1, X = N, P2 = 1;
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &x), S += x;
    if (S > T) break;
    P2 = (long long)P2 * Inv2 % Mod;
    Now = Now - (long long)Binom(i - 1, X) * P2 % Mod;
    if (Now < 0) Now += Mod;
    int Y = T - S < N ? T - S : N;
    while (X > Y) {
      Now = Now - (long long)Binom(i, X) * P2 % Mod;
      if (Now < 0) Now += Mod;
      --X;
    }
    Ans += Now;
    if (Ans >= Mod) Ans -= Mod;
  }
  printf("%d\n", Ans);
  return 0;
}
