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>
#include <iostream>
#include <string>
#include <cstring>
#include <set>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <algorithm>
#include <typeinfo>
#include <iomanip>
#include <cmath>
#include <cstdio>
//////  #include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<bool> vb;
typedef vector<long long> vll;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;

#define INF 1'000'000'000
#define mod 1'000'000'007
#define EPSILON 1e-9
#define MAXN 100001

int pre[101];
int shrt[101];
pii ans[101];
void solve() {
	int n, k;
	cin >> n >> k;
	memset(pre, 0, sizeof(pre));
	for (int i = 0; i < 101; ++i)
		shrt[i] = INF;
	for (int i = 1; i <= n * k; ++i) {
		int a;
		cin >> a;
		if (pre[a] == 0)
			pre[a] = i;
		else {
			if (shrt[a] > (i - pre[a])) {
				shrt[a] = i - pre[a];
				ans[a] = { pre[a],i };
			}
			pre[a] = i;
		}
	}
	for (int i = 1; i <= n; ++i) {
		cout << ans[i].first << " " << ans[i].second << "\n";
	}
}

int main()
{
	cin.sync_with_stdio(false);
	cin.tie(0);
	cout.sync_with_stdio(false);
	cout.tie(0);

	solve();

	return 0;
}
