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;
const int N = 2e5 + 5, P = 1e9 + 7;
int n, fc[N], ifc[N], p[N], ans, imc2;
long long I[N];
int inv(int x) { return x == 1 ? x : 1LL * (P - P / x) * inv(P % x) % P; }
int C(int n, long long m) {
  return n < m ? 0 : 1LL * fc[n] * ifc[m] % P * ifc[n - m] % P;
}
int main() {
  scanf("%d%lld", &n, I);
  for (int i = 1, x; i <= n; ++i) scanf("%d", &x), I[i] = I[i - 1] - x;
  fc[0] = ifc[0] = 1;
  for (int i = 1; i <= n; ++i) fc[i] = 1LL * i * fc[i - 1] % P;
  ifc[n] = inv(fc[n]), imc2 = (P + 1) >> 1;
  for (int i = n - 1; i; --i) ifc[i] = 1LL * (i + 1) * ifc[i + 1] % P;
  for (int i = 0; i <= min(1LL, I[1]); ++i) (p[1] += C(1, i)) %= P;
  (ans += 1LL * p[1] * imc2 % P) %= P;
  for (int i = 2; i <= n; ++i) {
    if (I[i] < 0) break;
    imc2 = 1LL * imc2 * (P + 1 >> 1) % P;
    p[i] = (2 * p[i - 1] % P - C(i - 1, I[i - 1]) + P) % P;
    for (int j = min<long long>(i, I[i - 1]); j >= I[i] + 1; --j)
      p[i] = (p[i] - C(i, j) + P) % P;
    (ans += 1LL * p[i] * imc2 % P) %= P;
  }
  printf("%d", ans);
  return 0;
}
