The numbers 1,   2,   ...,   n ⋅ k are colored with n colors. These colors are indexed by 1,   2,   ...,   n. For each 1 ≤ i ≤ n, there are exactly k numbers colored with color i.

Let [a,   b] denote the interval of integers between a and b inclusive, that is, the set \\{a,   a + 1,   ...,   b\}. You must choose n intervals [a_1,   b_1],   [a_2,   b_2],   ..., [a_n,   b_n] such that: 

  * for each 1 ≤ i ≤ n, it holds 1 ≤ a_i < b_i ≤ n ⋅ k; 
  * for each 1 ≤ i ≤ n, the numbers a_i and b_i are colored with color i; 
  * each number 1 ≤ x ≤ n ⋅ k belongs to at most \left⌈ (n)/(k - 1) \right⌉ intervals. 



One can show that such a family of intervals always exists under the given constraints.

Input

The first line contains two integers n and k (1 ≤ n ≤ 100, 2 ≤ k ≤ 100) — the number of colors and the number of occurrences of each color.

The second line contains n ⋅ k integers c_1,   c_2,   ...,   c_{nk} (1 ≤ c_j ≤ n), where c_j is the color of number j. It is guaranteed that, for each 1 ≤ i ≤ n, it holds c_j = i for exactly k distinct indices j.

Output

Output n lines. The i-th line should contain the two integers a_i and b_i.

If there are multiple valid choices of the intervals, output any.

Examples

Input


4 3
2 4 3 1 1 4 2 3 2 1 3 4


Output


4 5
1 7
8 11
6 12

Input


1 2
1 1


Output


1 2


Input


3 3
3 1 2 3 2 1 2 1 3


Output


6 8
3 7
1 4

Input


2 3
2 1 1 1 2 2


Output


2 3
5 6

Note

In the first sample, each number can be contained in at most \left⌈ (4)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:

<image>

In the second sample, the only interval to be chosen is forced to be [1,   2], and each number is indeed contained in at most \left⌈ (1)/(2 - 1) \right⌉ = 1 interval.

In the third sample, each number can be contained in at most \left⌈ (3)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:

<image>
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Math.max;
import static java.lang.System.exit;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class E {

	static void sortBy(int a[], int n, int v[]) {
		if (n == 0) {
			return;
		}
		for (int i = 1; i < n; i++) {
			int j = i;
			int ca = a[i];
			int cv = v[ca];
			do {
				int nj = (j - 1) >> 1;
				int na = a[nj];
				if (cv <= v[na]) {
					break;
				}
				a[j] = na;
				j = nj;
			} while (j != 0);
			a[j] = ca;
		}
		int ca = a[0];
		for (int i = n - 1; i > 0; i--) {
			int j = 0;
			while ((j << 1) + 2 + Integer.MIN_VALUE < i + Integer.MIN_VALUE) {
				j <<= 1;
				j += (v[a[j + 2]] > v[a[j + 1]]) ? 2 : 1;
			}
			if ((j << 1) + 2 == i) {
				j = (j << 1) + 1;
			}
			int na = a[i];
			a[i] = ca;
			ca = na;
			int cv = v[ca];
			while (j != 0 && v[a[j]] < cv) {
				j = (j - 1) >> 1;
			}
			while (j != 0) {
				na = a[j];
				a[j] = ca;
				ca = na;
				j = (j - 1) >> 1;
			}
		}
		a[0] = ca;
	}

	static void solve() throws Exception {
		int n = scanInt(), k = scanInt(), cnt[] = new int[n], occ[][] = new int[k][n];
		for (int i = 0; i < n * k; i++) {
			int c = scanInt() - 1;
			occ[cnt[c]++][c] = -i;
		}
		int idx[] = new int[n];
		for (int i = 0; i < n; i++) {
			idx[i] = i;
		}
		int p = n, s = (n + k - 2) / (k - 1);
		int a[] = new int[n], b[] = new int[n];
		for (int i = 1; i < k; i++) {
			sortBy(idx, p, occ[i]);
			int np = max(p - s, 0);
			for (int j = np; j < p; j++) {
				int jj = idx[j];
				a[jj] = -occ[i - 1][jj];
				b[jj] = -occ[i][jj];
			}
			p = np;
		}
		for (int i = 0; i < n; i++) {
			out.println((a[i] + 1) + " " + (b[i] + 1));
		}
	}

	static int scanInt() throws IOException {
		return parseInt(scanString());
	}

	static long scanLong() throws IOException {
		return parseLong(scanString());
	}

	static String scanString() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	public static void main(String[] args) {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			solve();
			in.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}