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>
const int mod = 1000000007;
int add32(int x, int y) {
  int res = x + y;
  if (res >= mod) res -= mod;
  return res;
}
int sub32(int x, int y) {
  int res = x - y;
  if (res < 0) res += mod;
  return res;
}
int mul32(int x, int y) { return (int64_t(x) * y) % mod; }
template <class T>
std::pair<T, T> gcd2(T x, T y) {
  if (x == 0) {
    return {0, 1};
  }
  T s = y / x;
  std::pair<T, T> res = gcd2(y % x, x);
  return {res.second - s * res.first, res.first};
}
template <class T>
T inv(T x) {
  x = x % mod;
  if (x < 0) {
    x += mod;
  }
  std::pair<T, T> res = gcd2(x, mod);
  if (res.first < 0) {
    res.first += mod;
  }
  if (res.first >= mod) {
    res.first -= mod;
  }
  return res.first;
}
struct Solution {
  int n;
  int64_t T;
  std::vector<int64_t> ts;
  std::vector<int> pow2;
  std::vector<int> ipow2;
  std::vector<int> fact;
  std::vector<int> ifact;
  int inv2;
  void precalc() {
    pow2.resize(n + 1);
    pow2[0] = 1;
    for (int i = 1; i < n; i++) {
      pow2[i] = mul32(pow2[i - 1], 2);
    }
    ipow2.resize(n + 1);
    ipow2[0] = 1;
    inv2 = inv(2);
    for (int i = 1; i < n; i++) {
      ipow2[i] = mul32(ipow2[i - 1], inv2);
    }
    fact.resize(n + 1);
    ifact.resize(n + 1);
    fact[0] = 1;
    ifact[0] = 1;
    for (int i = 1; i < n; i++) {
      fact[i] = mul32(fact[i - 1], i);
      ifact[i] = inv(fact[i]);
    }
  }
  int binProb(int k, int n) {
    return mul32(ipow2[n], mul32(fact[n], mul32(ifact[k], ifact[n - k])));
  }
  void run(std::istream &in, std::ostream &out) {
    in >> n >> T;
    ts.resize(n);
    for (int i = 0; i < n; i++) {
      in >> ts[i];
    }
    ts.push_back(T + 1);
    n++;
    precalc();
    int64_t sum = 0;
    int before = 1;
    int ans = 0;
    for (int i = 0; i < n && before > 0; i++) {
      int64_t sumNext = sum + ts[i];
      int next = (int)std::min<int64_t>(before + 1, T - sumNext + 1);
      if (next <= before) {
        int prob = next > 0 ? mul32(inv2, binProb(next - 1, i)) : 0;
        for (int j = std::max(next, 0); j < before; j++) {
          prob = add32(prob, binProb(j, i));
        }
        ans = add32(ans, mul32(prob, i));
      }
      before = next;
      sum = sumNext;
    }
    out << ans << "\n";
  }
};
int main() {
  std::cin.sync_with_stdio(false);
  std::cin.tie(nullptr);
  Solution().run(std::cin, std::cout);
  return 0;
}
