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;
using pii = pair<int, int>;
using ll = long long;
using pll = pair<ll, ll>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using ld = long double;
using ui = unsigned int;
using ull = unsigned long long;
using ui128 = __uint128_t;
const int MN = 2e5 + 2, LOG = 17, INF = 1e8, MOD = 1e9 + 7;
ll psa[MN];
int fac[MN], invfac[MN], pw2[MN];
int inv(int base) {
  int exp = MOD - 2, res = 1;
  while (exp) {
    if (exp & 1) res = res * 1LL * base % MOD;
    base = base * 1LL * base % MOD;
    exp >>= 1;
  }
  return res;
}
int choose(int n, int k) {
  return fac[n] * 1LL * invfac[k] % MOD * invfac[n - k] % MOD;
}
int main() {
  int n, least = 0, most = 0, tot = 0;
  ll t;
  scanf("%d %lld", &n, &t);
  fac[0] = invfac[0] = pw2[0] = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &psa[i]);
    psa[i] += psa[i - 1];
    fac[i] = fac[i - 1] * 1LL * i % MOD;
    pw2[i] = pw2[i - 1] * 2 % MOD;
    if (psa[i] + i <= t) least = i;
    if (psa[i] <= t) most = i;
  }
  invfac[n] = inv(fac[n]);
  for (int i = n - 1; i >= 1; i--)
    invfac[i] = invfac[i + 1] * 1LL * (i + 1) % MOD;
  for (int i = least; i <= most; i++) {
    int lx = i == n ? 0 : t - psa[i + 1] + 1, rx = t - psa[i];
    lx = max(lx, 0);
    rx = min(rx, i);
    for (int x = lx; x <= rx; x++) {
      tot += choose(i, x) * 1LL * pw2[n - i] % MOD * 1LL * i % MOD;
      if (tot >= MOD) tot -= MOD;
    }
    if (lx && i != n)
      tot += choose(i, lx - 1) * 1LL * pw2[n - i - 1] % MOD * 1LL * i % MOD;
    if (tot >= MOD) tot -= MOD;
  }
  printf("%lld\n", tot * 1LL * inv(pw2[n]) % MOD);
  return 0;
}
