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 MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int ddx[] = {-1, -2, 1, -2, 2, -1, 2, 1}, ddy[] = {-2, -1, -2, 1, -1, 2, 1, 2};
long long fac[200001], inv[200001], t[200002], p[200002];
long long nCr(int n, int r) {
  if (r < 0 || r > n) return 0;
  long long c = fac[n];
  c = (c * inv[r]) % MOD;
  c = (c * inv[n - r]) % MOD;
  return c;
}
int main() {
  int n;
  long long T;
  scanf("%d%lld", &n, &T);
  fac[0] = inv[0] = 1;
  for (int(i) = (1); (i) <= (200000); (i) += (1))
    fac[i] = (fac[i - 1] * i) % MOD;
  inv[200000] = POW(fac[200000], MOD - 2);
  for (int(i) = (199999); (i) >= (1); (i) -= (1))
    inv[i] = (inv[i + 1] * (i + 1)) % MOD;
  p[0] = 1, p[1] = inv[2];
  for (int(i) = (2); (i) <= (200000); (i) += (1))
    p[i] = (p[i - 1] * p[1]) % MOD;
  long long ans = 0, S = 0;
  for (int(i) = (1); (i) <= (n); (i) += (1)) scanf("%lld", t + i);
  t[n + 1] = 987654321987654321;
  long long tmp = 0, l = 0, r = -1;
  for (int(x) = (1); (x) <= (n); (x) += (1)) {
    S += t[x];
    long long L = max(0LL, T - S - t[x + 1] + 1), R = min(1LL * x, T - S);
    if (L <= R) {
      while (r < R) tmp = (tmp + nCr(x, ++r)) % MOD;
      while (r > R) tmp = (tmp + MOD - nCr(x, r--)) % MOD;
      while (l > L) tmp = (tmp + nCr(x, --l)) % MOD;
      while (l < L) tmp = (tmp + MOD - nCr(x, l++)) % MOD;
      ans = (ans + (tmp * p[x]) % MOD * x) % MOD;
    }
    long long LL = max(0LL, T - S - t[x + 1]);
    if (LL <= R) {
      while (r < R) tmp = (tmp + nCr(x, ++r)) % MOD;
      while (r > R) tmp = (tmp + MOD - nCr(x, r--)) % MOD;
      while (l > LL) tmp = (tmp + nCr(x, --l)) % MOD;
      while (l < LL) tmp = (tmp + MOD - nCr(x, l++)) % MOD;
      ans = (ans + (tmp * p[x]) % MOD * x) % MOD;
    }
    tmp = (tmp + tmp + nCr(x, l - 1) - nCr(x, r)) % MOD;
  }
  ans = (ans * inv[2]) % MOD;
  printf("%lld", ans);
}
