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 + 10;
const int mod = 1e9 + 7;
int addmod(int x, int y) {
  if (y < 0)
    x += y;
  else
    x = x - mod + y;
  if (x < 0) x += mod;
  return x;
}
int qpow(int a, int b, int mo) {
  int res = 1;
  while (b) {
    if (b & 1) res = 1ll * res * a % mo;
    a = 1ll * a * a % mo;
    b >>= 1;
  }
  return res;
}
int invfac2[N], invfac[N], fac[N];
void init() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  invfac[N - 1] = qpow(fac[N - 1], mod - 2, mod);
  invfac[0] = 1;
  for (int i = N - 2; i >= 1; i--)
    invfac[i] = 1ll * invfac[i + 1] * (i + 1) % mod;
  invfac2[1] = qpow(2, mod - 2, mod);
  for (int i = 2; i < N; i++)
    invfac2[i] = 1ll * invfac2[i - 1] * invfac2[1] % mod;
}
int Calc(int n, int m) {
  if (n > m)
    return 0;
  else
    return 1ll * fac[m] * invfac[n] % mod * invfac[m - n] % mod;
}
int n, ans;
long long T, t[N];
bool Flag = 0;
int main() {
  init();
  scanf("%d%lld", &n, &T);
  for (int i = 1; i <= n; i++) scanf("%lld", &t[i]);
  long long sum = 0;
  int s = 2;
  int r = n;
  for (int i = 1; i <= n; i++) {
    sum += t[i];
    if (T - sum >= i)
      ans = addmod(1, ans);
    else if (T - sum < 0)
      break;
    else {
      for (int j = T - sum + 1; j <= r; j++) s = addmod(s, -Calc(j, i));
      ans = addmod(1ll * s * invfac2[i] % mod, ans);
      r = T - sum;
      Flag = 1;
    }
    s = 2ll * s % mod;
    ;
    if (Flag) s = addmod(s, -Calc(r, i));
  }
  printf("%d\n", ans);
}
