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 long long mod = 1e9 + 7;
long long Pow(long long a, long long b) {
  long long x = 1;
  for (; b > 0; b >>= 1) {
    if (b & 1) x = x * a % mod;
    a = a * a % mod;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long n, t;
  cin >> n >> t;
  vector<long long> a;
  long long sum = 0;
  for (int i = 0, x; i < n; ++i) {
    cin >> x;
    sum += x;
    if (sum > t) break;
    a.push_back(x);
  }
  n = a.size();
  vector<long long> s(n + 1);
  for (int i = 0; i < n; ++i) s[i + 1] = s[i] + a[i];
  if (t - s[n] >= n) {
    cout << n << "\n";
    return 0;
  }
  vector<long long> fac(n + 1), ifac(n + 1);
  fac[0] = ifac[0] = 1;
  for (int i = 1; i <= n; ++i) {
    fac[i] = fac[i - 1] * i % mod;
    ifac[i] = Pow(fac[i], mod - 2);
  }
  auto comb = [&](long long i, long long j) {
    return fac[i] * ifac[i - j] % mod * ifac[j] % mod;
  };
  s.push_back(t + 1);
  long long f1 = -1, f2, ans = 0, g, R = sqrt(n) + 1, p2 = 1;
  for (int i = 1; i <= n; ++i) {
    p2 = p2 * 2 % mod;
    long long k2 = max(0LL, t - s[i + 1]), k = max(0LL, t - s[i + 1] + 1),
              to = min((long long)i, t - s[i]);
    if (k2 != k && k2 <= to) {
      ans = (ans + comb(i, k2) * Pow(p2 * 2 % mod, mod - 2) % mod * i) % mod;
    }
    if (k > to) continue;
    if (f1 == -1 || f1 - k > R || f2 - to > R) {
      g = 0;
      for (int j = k; j <= to; ++j) g = (g + comb(i, j)) % mod;
    } else {
      g = (g * 2 + mod - comb(i - 1, f2) +
           ((f1 > 0) ? comb(i - 1, f1 - 1) : 0)) %
          mod;
      while (f2 < to) g = (g + comb(i, ++f2)) % mod;
      while (f2 > to) g = (g + mod - comb(i, f2--)) % mod;
      while (f1 > k) g = (g + comb(i, --f1)) % mod;
    }
    ans = (ans + g * Pow(p2, mod - 2) % mod * i) % mod;
    f1 = k;
    f2 = to;
  }
  cout << ans << "\n";
  return 0;
}
