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 lint = long long int;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios {
  fast_ios() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  };
} fast_ios_;
template <typename T>
void ndarray(vector<T> &vec, int len) {
  vec.resize(len);
}
template <typename T, typename... Args>
void ndarray(vector<T> &vec, int len, Args... args) {
  vec.resize(len);
  for (auto &v : vec) ndarray(v, args...);
}
template <typename T>
bool mmax(T &m, const T q) {
  if (m < q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T>
bool mmin(T &m, const T q) {
  if (m > q) {
    m = q;
    return true;
  } else
    return false;
}
template <typename T1, typename T2>
pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) {
  return make_pair(l.first + r.first, l.second + r.second);
}
template <typename T1, typename T2>
pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) {
  return make_pair(l.first - r.first, l.second - r.second);
}
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
  for (auto &v : vec) is >> v;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << "[";
  for (auto v : vec) os << v << ",";
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const deque<T> &vec) {
  os << "deq[";
  for (auto v : vec) os << v << ",";
  os << "]";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &vec) {
  os << "{";
  for (auto v : vec) os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &vec) {
  os << "{";
  for (auto v : vec) os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &vec) {
  os << "{";
  for (auto v : vec) os << v << ",";
  os << "}";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) {
  os << "{";
  for (auto v : vec) os << v << ",";
  os << "}";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &pa) {
  os << "(" << pa.first << "," << pa.second << ")";
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const map<TK, TV> &mp) {
  os << "{";
  for (auto v : mp) os << v.first << "=>" << v.second << ",";
  os << "}";
  return os;
}
template <typename TK, typename TV>
ostream &operator<<(ostream &os, const unordered_map<TK, TV> &mp) {
  os << "{";
  for (auto v : mp) os << v.first << "=>" << v.second << ",";
  os << "}";
  return os;
}
constexpr lint MOD = 1000000007;
lint N, T;
vector<lint> fac, facInv, inv;
void facInit(int nmax) {
  fac = facInv = inv = vector<lint>(nmax + 1, 1);
  for (int i = 2; i <= nmax; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    facInv[i] = facInv[i - 1] * inv[i] % MOD;
  }
}
lint nCr(int n, int r) {
  if (n < r || r < 0) return 0;
  if (n >= (int)fac.size()) facInit(n);
  return (fac[n] * facInv[r] % MOD) * facInv[n - r] % MOD;
}
lint nPr(int n, int r) {
  if (n < r || r < 0) return 0;
  if (n >= (int)fac.size()) facInit(n);
  return fac[n] * facInv[n - r] % MOD;
}
lint power(lint x, lint n, lint mod = MOD) {
  lint ans = 1;
  while (n > 0) {
    if (n & 1) (ans *= x) %= mod;
    (x *= x) %= mod;
    n >>= 1;
  }
  return ans;
}
lint extgcd(lint a, lint b, lint &x, lint &y) {
  lint d = a;
  if (b != 0)
    d = extgcd(b, a % b, y, x), y -= (a / b) * x;
  else
    x = 1, y = 0;
  return d;
}
lint mod_inverse(lint a, lint m = MOD) {
  lint x, y;
  extgcd(a, m, x, y);
  return (m + x % m) % m;
}
int main() {
  cin >> N >> T;
  vector<lint> t(N);
  for (int i = (0), i_end_ = (N); i < i_end_; i++) cin >> t[i];
  t.push_back(300000000000000LL);
  N++;
  facInit(N);
  lint ret = 0;
  lint acc_time = 0;
  int i = 0;
  lint prb = 0;
  while (i < N and acc_time < T) acc_time += t[i], i++;
  for (; i > 0; i--) {
    if (acc_time + i <= T)
      prb = 1;
    else if (acc_time > T)
      prb = 0;
    else {
      lint d = T - acc_time;
      lint mip = mod_inverse(power(2, i + 1));
      for (int j = (max<lint>(0, d - t[i] + 1)), j_end_ = (d + 1); j < j_end_;
           j++)
        prb += nCr(i, j) * mip % MOD;
      for (int j = (max<lint>(0, d - (t[i] + 1) + 1)), j_end_ = (d + 1);
           j < j_end_; j++)
        prb += nCr(i, j) * mip % MOD;
    }
    acc_time -= t[i - 1];
    ret += prb;
  }
  cout << ret % MOD << endl;
}
