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>
template <std::uint32_t mod>
class modint {
 private:
  std::uint32_t n;

 public:
  modint() : n(0){};
  modint(std::int64_t n_) : n((n_ >= 0 ? n_ : mod - (-n_) % mod) % mod){};
  static constexpr std::uint32_t get_mod() { return mod; }
  std::uint32_t get() const { return n; }
  bool operator==(const modint& m) const { return n == m.n; }
  bool operator!=(const modint& m) const { return n != m.n; }
  modint& operator+=(const modint& m) {
    n += m.n;
    n = (n < mod ? n : n - mod);
    return *this;
  }
  modint& operator-=(const modint& m) {
    n += mod - m.n;
    n = (n < mod ? n : n - mod);
    return *this;
  }
  modint& operator*=(const modint& m) {
    n = std::uint64_t(n) * m.n % mod;
    return *this;
  }
  modint operator+(const modint& m) const { return modint(*this) += m; }
  modint operator-(const modint& m) const { return modint(*this) -= m; }
  modint operator*(const modint& m) const { return modint(*this) *= m; }
  modint inv() const { return (*this).pow(mod - 2); }
  modint pow(std::uint64_t b) const {
    modint ans = 1, m = modint(*this);
    while (b) {
      if (b & 1) ans *= m;
      m *= m;
      b >>= 1;
    }
    return ans;
  }
};
using namespace std;
using modulo = modint<1000000007>;
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int N;
  long long T;
  cin >> N >> T;
  vector<modulo> fact(N + 1), inv(N + 1), factinv(N + 1);
  fact[0] = 1;
  inv[1] = 1;
  factinv[0] = 1;
  for (int i = 1; i <= N; ++i) fact[i] = fact[i - 1] * i;
  for (int i = 2; i <= N; ++i)
    inv[i] = inv[modulo::get_mod() % i] *
             (modulo(0) - modulo(modulo::get_mod() / i));
  for (int i = 1; i <= N; ++i) factinv[i] = factinv[i - 1] * inv[i];
  function<modulo(int, int)> comb = [&](int x, int y) {
    if (y < 0 || x < y) return modulo(0);
    return fact[x] * factinv[y] * factinv[x - y];
  };
  int curr = 0;
  modulo d = 1, ex = 0, inv2 = modulo(2).inv(), pw = 1;
  for (int i = 1; i <= N; ++i) {
    long long x;
    cin >> x;
    T -= x;
    if (T < 0) break;
    int nxtr = min(T, (long long)(i));
    d *= 2;
    if (curr + 1 > nxtr) {
      d -= comb(i - 1, curr);
      for (int j = curr; j > nxtr; --j) {
        d -= comb(i - 1, j) + comb(i - 1, j - 1);
      }
    }
    pw *= inv2;
    ex += d * pw;
    curr = nxtr;
  }
  cout << ex.get() << endl;
  return 0;
}
