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>
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
char nextChar() {
  char x;
  scanf("%c", &x);
  return x;
}
void newLine() { printf("\n"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long pow(long long a, long long b, long long MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
int countSetBit(long long n) {
  int ans = 0;
  while (n != 0) {
    n -= (n & -n);
    ++ans;
  }
  return ans;
}
long long mod = 1ll * 1e9 + 7;
const int N = 3010;
const int M = 6;
const double eps = 1e-6;
using namespace std;
int a[N], b[N];
vector<pair<int, int> > v;
int main() {
  int n = nextInt(), k = nextInt();
  if (k > n * n) {
    puts("-1");
    return 0;
  }
  int a[101][101];
  memset(a, 0, sizeof a);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      if (k == 0) {
        break;
      }
      if (i == j) {
        if ((k - 1) % 2 == 0 || ((k - 2) % 2 == 0 && i + 1 < n)) {
          a[i][j] = 1;
          k -= 1;
        }
      }
      if (k > 1 && i != j) {
        a[i][j] = a[j][i] = 1;
        k -= 2;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << a[i][j] << " ";
    }
    newLine();
  }
  return 0;
}