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 = 1e9 + 7;
const int inv2 = (mod + 1) >> 1;
int n, curn, curk;
long long T, sigmaC, E, inv2x;
const int MAXN = 2e5;
long long inv[MAXN + 5], fac[MAXN + 5], invfac[MAXN + 5];
void init_inv(int n = MAXN) {
  inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  }
}
void init_fac_invfac(int n = MAXN) {
  init_inv(n);
  fac[0] = 1, invfac[0] = 1;
  for (int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % mod;
    invfac[i] = invfac[i - 1] * inv[i] % mod;
  }
}
inline long long C(long long n, long long m) {
  if (n < m) return 0;
  return fac[n] * invfac[n - m] % mod * invfac[m] % mod;
}
int t[200005];
long long pret[200005];
void calc_sigmaC(int x) {
  long long tmp = 0;
  int dx = min((long long)x, T - pret[x]);
  for (int i = 0; i <= dx; i++) {
    tmp += C(x, i);
    if (tmp >= mod) tmp -= mod;
  }
  sigmaC = tmp;
  curn = x;
  curk = dx;
}
void next_sigmaC(int nextx) {
  int dx1 = min((long long)nextx, T - pret[nextx]);
  long long tmp = sigmaC;
  tmp = 2ll * tmp % mod;
  tmp = (tmp + C(curn, curk + 1)) % mod;
  curn++;
  curk++;
  while (curk > dx1) {
    tmp = (tmp + mod - C(curn, curk)) % mod;
    curk--;
  }
  sigmaC = tmp;
}
int main() {
  init_fac_invfac();
  while (~scanf("%d%lld", &n, &T)) {
    pret[0] = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &t[i]);
      pret[i] = pret[i - 1] + t[i];
    }
    E = 0;
    inv2x = inv2;
    int x = 1;
    while (x <= n && pret[x] + x <= T) {
      x++;
      E++;
      inv2x = (inv2x * inv2) % mod;
    }
    if (x <= n) {
      calc_sigmaC(x);
      for (; x <= n; x++) {
        E = (E + inv2x * sigmaC % mod) % mod;
        if (T - pret[x + 1] < 0) break;
        next_sigmaC(x + 1);
        inv2x = (inv2x * inv2) % mod;
      }
    }
    printf("%lld\n", E);
  }
}
