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;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
int mod = 1000000007;
inline int mul(int x, int y) { return 1ll * x * y % mod; }
int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int sub(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
int sq(int x) { return 1ll * x * x % mod; }
int pow(int a, int b) {
  return b == 0 ? 1 : (b & 1 ? mul(a, sq(pow(a, b / 2))) : sq(pow(a, b / 2)));
}
int n, t[1000020], fac[1000020], invfac[1000020];
int C(int n, int m) { return mul(fac[n], mul(invfac[n - m], invfac[m])); }
int cv = 1;
long long T, lwb = 0, upb = 0, mb = 0;
int main() {
  read(n);
  read(T);
  fac[0] = 1;
  for (int i = 1; i <= 1000010; i++) fac[i] = mul(i, fac[i - 1]);
  invfac[1000010] = pow(fac[1000010], mod - 2);
  for (int i = 1000010 - 1; i >= 0; i--) invfac[i] = mul(i + 1, invfac[i + 1]);
  int inv2 = (mod + 1) / 2, accu = 1, ans = 0;
  for (int i = 0; i < n; i++) {
    read(t[i]);
    lwb += t[i];
    upb += t[i] + 1;
    mb += t[i];
    if (lwb > T) break;
    cv = sub(mul(cv, 2), C(upb - lwb - 1, mb - lwb));
    while (mb < T && mb < upb) {
      cv = add(cv, C(upb - lwb, mb - lwb + 1));
      mb += 1;
    }
    while (mb > T) {
      cv = sub(cv, C(upb - lwb, mb - lwb));
      mb -= 1;
    }
    accu = mul(accu, inv2);
    ans = add(ans, mul(cv, accu));
  }
  cout << ans << endl;
  return 0;
}
