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 = 200005;
const long long mod = 1e9 + 7;
const long long INF = 1e15;
int n;
long long T, fac[N], inv[N], binpw[N], t[N], P[N];
void add(long long &x, long long y) {
  x = (x + y) % mod;
  if (x < 0) x += mod;
}
void mul(long long &x, long long y) { x = x * y % mod; }
long long pw(long long x, long long p) {
  long long ret = 1, base = x;
  while (p) {
    if (p & 1) mul(ret, base);
    mul(base, base);
    p >>= 1;
  }
  return ret;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % mod;
  inv[n] = pw(fac[n], mod - 2);
  for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mod;
  binpw[0] = 1;
  for (int i = 1; i <= n; i++) binpw[i] = 2 * binpw[i - 1] % mod;
}
long long C(long long x, long long y) {
  if (y < 0 || x < y) return 0LL;
  return fac[x] * inv[y] % mod * inv[x - y] % mod;
}
int main() {
  scanf("%d%lld", &n, &T);
  init();
  for (int i = 1; i <= n; i++) scanf("%lld", &t[i]), P[i] = P[i - 1] + t[i];
  t[n + 1] = P[n + 1] = INF;
  int l = n + 1, r = n;
  long long sum_C = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    long long L = max(0LL, T - P[i + 1] + 1);
    long long R = min((long long)n, T - P[i]);
    long long p = 0;
    if (L <= R) {
      while (l && l > L) add(sum_C, C(i, --l));
      while (r && r > R) add(sum_C, -C(i, r--));
      add(p, sum_C * pw(binpw[i], mod - 2) % mod);
    }
    long long sum_C_2 = sum_C;
    add(sum_C_2, -C(i, r));
    add(sum_C_2, C(i, l - 1));
    add(sum_C, sum_C_2);
    if (i < n) add(p, C(i, T - P[i + 1]) * pw(binpw[i + 1], mod - 2) % mod);
    add(ans, i * p % mod);
  }
  cout << ans << endl;
}
