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 namespace std;
template <typename T, typename U>
std::istream& operator>>(std::istream& i, pair<T, U>& p) {
  i >> p.first >> p.second;
  return i;
}
template <typename T>
std::istream& operator>>(std::istream& i, vector<T>& t) {
  for (auto& v : t) {
    i >> v;
  }
  return i;
}
template <typename T, typename U>
std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {
  o << p.first << ' ' << p.second;
  return o;
}
template <typename T>
std::ostream& operator<<(std::ostream& o, const vector<T>& t) {
  if (t.empty()) o << '\n';
  for (size_t i = 0; i < t.size(); ++i) {
    o << t[i] << " \n"[i == t.size() - 1];
  }
  return o;
}
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T, vector<T>, less<T>>;
template <typename T>
bool in(T a, T b, T c) {
  return a <= b && b < c;
}
unsigned int logceil(int x) {
  return (unsigned int)(8 * sizeof(int) - __builtin_clz(x));
}
namespace std {
template <typename T, typename U>
struct hash<pair<T, U>> {
  hash<T> t;
  hash<U> u;
  size_t operator()(const pair<T, U>& p) const {
    return t(p.x) ^ (u(p.y) << 7);
  }
};
}  // namespace std
template <typename T>
T gcd(T a, T b) {
  if (a < b) swap(a, b);
  return b ? gcd(b, a % b) : a;
}
template <typename T>
class vector2 : public vector<vector<T>> {
 public:
  vector2() {}
  vector2(size_t a, size_t b, T t = T())
      : vector<vector<T>>(a, vector<T>(b, t)) {}
};
template <typename T>
class vector3 : public vector<vector2<T>> {
 public:
  vector3() {}
  vector3(size_t a, size_t b, size_t c, T t = T())
      : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}
};
template <typename T>
class vector4 : public vector<vector3<T>> {
 public:
  vector4() {}
  vector4(size_t a, size_t b, size_t c, size_t d, T t = T())
      : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}
};
template <typename T>
class vector5 : public vector<vector4<T>> {
 public:
  vector5() {}
  vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())
      : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}
};
void fastIO() { ios::sync_with_stdio(false); }
template <typename T, typename... Args>
T max(T t, Args... args) {
  return max(max<T>(args...), t);
}
template <typename T, typename... Args>
T min(T t, Args... args) {
  return min(min<T>(args...), t);
}
template <unsigned int N>
class Field {
  inline unsigned int pow(unsigned int a, unsigned int p) {
    unsigned int r = 1, e = a;
    while (p) {
      if (p & 1) {
        r = ((unsigned long long)r * e) % N;
      }
      e = ((unsigned long long)e * e) % N;
      p >>= 1;
    }
    return r;
  }
  inline unsigned int inv(unsigned int a) { return pow(a, N - 2); }

 public:
  inline Field(long long x = 0) : v((unsigned int)(x % N)) {}
  inline Field<N> pow(int p) { return (*this) ^ p; }
  inline Field<N> operator^(int p) { return {(int)pow(v, (unsigned int)p)}; }
  inline Field<N>& operator+=(const Field<N>& o) {
    if ((long long)v + o.v >= N)
      v += o.v - N;
    else
      v += o.v;
    return *this;
  }
  inline Field<N>& operator-=(const Field<N>& o) {
    if (v < o.v)
      v -= o.v - N;
    else
      v -= o.v;
    return *this;
  }
  inline Field<N>& operator*=(const Field<N>& o) {
    v = (unsigned int)((unsigned long long)v * o.v % N);
    return *this;
  }
  inline Field<N>& operator/=(const Field<N>& o) { return *this *= inv(o.v); }
  inline Field<N> operator+(const Field<N>& o) const {
    Field<N> r{*this};
    return r += o;
  }
  inline Field<N> operator-(const Field<N>& o) const {
    Field<N> r{*this};
    return r -= o;
  }
  inline Field<N> operator*(const Field<N>& o) const {
    Field<N> r{*this};
    return r *= o;
  }
  inline Field<N> operator/(const Field<N>& o) const {
    Field<N> r{*this};
    return r /= o;
  }
  inline Field<N> operator-() {
    if (v)
      return {(int)(N - v)};
    else
      return {0};
  };
  inline Field<N>& operator++() {
    ++v;
    if (v == N) v = 0;
    return *this;
  }
  inline Field<N> operator++(int) {
    Field<N> r{*this};
    ++*this;
    return r;
  }
  inline Field<N>& operator--() {
    --v;
    if (v == -1) v = N - 1;
    return *this;
  }
  inline Field<N> operator--(int) {
    Field<N> r{*this};
    --*this;
    return r;
  }
  inline bool operator==(const Field<N>& o) const { return o.v == v; }
  inline bool operator!=(const Field<N>& o) const { return o.v != v; }
  inline explicit operator unsigned int() const { return v; }
  inline static vector<Field<N>> fact(int t) {
    vector<Field<N>> F(t + 1, 1);
    for (int i = 2; i <= t; ++i) {
      F[i] = F[i - 1] * i;
    }
    return F;
  }
  inline static vector<Field<N>> invfact(int t) {
    vector<Field<N>> F(t + 1, 1);
    Field<N> first{1};
    for (int i = 2; i <= t; ++i) {
      first = first * i;
    }
    F[t] = 1 / first;
    for (int i = t - 1; i >= 2; --i) {
      F[i] = F[i + 1] * (i + 1);
    }
    return F;
  }

 private:
  unsigned int v;
};
template <unsigned int N>
istream& operator>>(std::istream& is, Field<N>& f) {
  unsigned int v;
  is >> v;
  f = v;
  return is;
}
template <unsigned int N>
ostream& operator<<(std::ostream& os, const Field<N>& f) {
  return os << (unsigned int)f;
}
template <unsigned int N>
Field<N> operator+(int i, const Field<N>& f) {
  return Field<N>(i) + f;
}
template <unsigned int N>
Field<N> operator-(int i, const Field<N>& f) {
  return Field<N>(i) - f;
}
template <unsigned int N>
Field<N> operator*(int i, const Field<N>& f) {
  return Field<N>(i) * f;
}
template <unsigned int N>
Field<N> operator/(int i, const Field<N>& f) {
  return Field<N>(i) / f;
}
struct Ring {
  static int div(int p, int q, int N) {
    long long t = 0, nt = 1, r = N, nr = q;
    while (nr) {
      long long qq = r / nr;
      t -= qq * nt;
      r -= qq * nr;
      swap(t, nt);
      swap(r, nr);
    }
    t = (t < 0) ? t + N : t;
    r = (r < 0) ? r + N : r;
    if (gcd(p, N) % r) {
      return 0;
    }
    return (int)(((long long)t * (long long)p / r) % N);
  }
};
const int MAXN = 200200;
auto fact = Field<1000000007>::fact(MAXN);
auto ifact = Field<1000000007>::invfact(MAXN);
Field<1000000007> C(int n, int k) { return fact[n] * ifact[k] * ifact[n - k]; }
class FEkspertPoKrossvordam {
 public:
  void solve(std::istream& in, std::ostream& out) {
    fastIO();
    int n;
    long long T;
    in >> n >> T;
    vector<long long> a(n);
    in >> a;
    Field<1000000007> ans = 0;
    Field<1000000007> cs = 1;
    long long k = 0;
    Field<1000000007> itwo = Field<1000000007>(1) / 2;
    long long ct = 0;
    for (int i = 1; i < n + 1; ++i) {
      ct += a[i - 1];
      if (ct > T) break;
      int nk = min(T - ct, (long long)i);
      Field<1000000007> tot = 0;
      if (k != 0) {
        cs = 2 * cs - C(i - 1, k);
      }
      while (k < nk) {
        cs += C(i, k + 1);
        ++k;
      }
      while (k > nk) {
        cs -= C(i, k);
        --k;
      }
      ans += cs * itwo.pow(i);
    }
    out << ans << endl;
  }
};
int main() {
  FEkspertPoKrossvordam solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
