Problem: You are given matrix with n rows and n columns filled with zeroes. You should put k ones in it in such a way that the resulting matrix is symmetrical with respect to the main diagonal (the diagonal that goes from the top left to the bottom right corner) and is lexicographically maximal.

One matrix is lexicographically greater than the other if the first different number in the first different row from the top in the first matrix is greater than the corresponding number in the second one.

If there exists no such matrix then output -1.

Input

The first line consists of two numbers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 106).

Output

If the answer exists then output resulting matrix. Otherwise output -1.

Examples

Input

2 1


Output

1 0 
0 0 


Input

3 2


Output

1 0 0 
0 1 0 
0 0 0 


Input

2 5


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void unique(T &c) {
  c.erase(std::unique(c.begin(), c.end()), c.end());
}
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
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) {
  for (int i = 0; i < vec.size(); i++) {
    os << vec[i];
    if (i + 1 != vec.size()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_set<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const unordered_multiset<T> &st) {
  for (auto itr = st.begin(); itr != st.end(); ++itr) {
    os << *itr;
    auto titr = itr;
    if (++titr != st.end()) os << " ";
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const map<T1, T2> &mp) {
  for (auto itr = mp.begin(); itr != mp.end(); ++itr) {
    os << "(" << itr->first << ", " << itr->second << ")";
    auto titr = itr;
    if (++titr != mp.end()) os << " ";
  }
  return os;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp) {
  for (auto itr = mp.begin(); itr != mp.end(); ++itr) {
    os << "(" << itr->first << ", " << itr->second << ")";
    auto titr = itr;
    if (++titr != mp.end()) os << " ";
  }
  return os;
}
using ll = long long int;
using P = pair<int, int>;
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
struct fast_io {
  fast_io() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(20);
  }
} fast_io_;
int res[110][110];
int main(int argc, char const *argv[]) {
  int n, k;
  cin >> n >> k;
  if (k > n * n) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = (0); i < (n); ++i) {
    if (k == 0) break;
    if (k % 2 == 1 || i < n - 1) {
      res[i][i] = 1;
      k--;
    }
    for (int j = i + 1; j < n; j++) {
      if (k >= 2) {
        res[i][j] = res[j][i] = 1;
        k -= 2;
      }
    }
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (n); ++j) {
      cout << res[i][j] << "\n "[j + 1 != n];
    }
  }
  cout << flush;
  return 0;
}