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 ll = long long;
template <class T, class U>
using Pa = pair<T, U>;
template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vec<T>>;
constexpr ll mod = 1e9 + 7;
struct mint {
  ll x;
  mint(ll x = 0) : x((x % mod + mod) % mod) {}
  friend ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }
  friend istream& operator>>(istream& is, mint& a) {
    ll t;
    is >> t;
    a = mint(t);
    return (is);
  }
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(ll t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  mint inv() const { return pow(mod - 2); }
  mint& operator/=(const mint a) { return (*this) *= a.inv(); }
  mint operator/(const mint a) const {
    mint res(*this);
    return res /= a;
  }
};
class combination {
 private:
  vector<mint> fact, inv, finv;

 public:
  combination(int N) {
    fact = inv = finv = vector<mint>(N + 1);
    fact[0] = fact[1] = 1;
    inv[0] = inv[1] = 1;
    finv[0] = finv[1] = 1;
    for (ll i = 2; i <= N; i++) {
      fact[i] = fact[i - 1] * i;
      inv[i] = (mint)mod - inv[mod % i] * (mod / i);
      finv[i] = finv[i - 1] * inv[i];
    }
  }
  mint f(int i) { return fact[i]; }
  mint comb(int n, int k) {
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fact[n] * finv[k] * finv[n - k];
  }
  mint hcomb(int n, int k) {
    if (n == 0 && k == 0) return 1;
    return comb(n + k - 1, k);
  }
};
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  combination C(N);
  ll T;
  cin >> T;
  mint ans = 0, two = 2, inv2 = two.inv();
  mint num = 1, prob = 1;
  ll sum = 0;
  ll now = -1;
  for (ll i = 1; i <= N; i++) {
    ll t;
    cin >> t;
    sum += t;
    ll l = T - sum;
    num *= two;
    num -= C.comb(i - 1, (now == -1 ? i : now));
    prob *= inv2;
    if (l < 0)
      continue;
    else if (i <= l)
      ans += 1;
    else {
      if (now == -1) now = i;
      while (now > l) {
        num -= C.comb(i, now);
        now--;
      }
      ans += num * prob;
    }
  }
  cout << ans << "\n";
}
