Problem: Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.

Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.

Input

The first line of input contains three space-separated integers n, k, d (1 ≤ n, d ≤ 1000; 1 ≤ k ≤ 109).

Output

If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.

Examples

Input

3 2 2


Output

1 1 2 
1 2 1 


Input

3 2 1


Output

-1

Note

Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long long max(long long a, long long b) { return ((a > b) ? a : b); }
inline long long min(long long a, long long b) { return ((a > b) ? b : a); }
inline long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long a = power(x, y / 2);
  if (y % 2 == 0)
    return a * a;
  else
    return x * a * a;
}
long long powermod(long long x, long long ex, long long md) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % md;
    ex >>= 1ll;
    x = (x * x) % md;
  }
  return ans;
}
const long long inf = 1e18 + 9;
const long long mod = 1e9 + 7;
const long double eps = 1e-9;
const long long N = 1e6 + 1;
long long n, k, d;
void solve() {
  bool flag = 0;
  cin >> n >> k >> d;
  long long arr[n + 1];
  long long p = 1;
  for (long long i = 1; i <= d; i++) {
    p = p * k;
    if (p >= n) {
      flag = 1;
      break;
    }
  }
  if (!flag)
    cout << "-1\n";
  else {
    for (int(i) = 0; i < (n); i++) arr[i + 1] = i + 1;
    for (long long i = 1; i <= d; i++) {
      for (long long j = 1; j <= n; j++) {
        cout << arr[j] % k + 1 << " ";
        arr[j] /= k;
      }
      cout << "\n";
    }
  }
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  cout << fixed << setprecision(20);
  long long NTC = 1;
  long long PTC = 0;
  while (PTC++ < NTC) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}