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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update
using namespace __gnu_pbds; using namespace std;
// containers
typedef long long ll;typedef double db; typedef long double ld;
typedef pair<int,int> pii;typedef pair<ll,ll> pll;typedef pair<db,db> pdb;
typedef vector<int>vi;typedef vector<ll>vll;typedef vector<pii>vpi;typedef vector<pll>vpl; typedef vector<db>vdb;
typedef vector<vi>vvi;typedef vector<vll>vvl;typedef vector<vpi>vvpi;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> oset;
#define tcT template<class T
#define tcTU tcT, class U
tcT>using vec=vector<T>;
tcT>using pq=priority_queue<T>;
tcT>using uset=unordered_set<T>;
tcTU>using umap=unordered_map<T,U>;
tcT,size_t SZ>using ar=array<T,SZ>;
// macros
#define siz(x) (int)x.size()
#define af(x) x.begin(), x.end()
#define ms(x, y) memset(x, y, sizeof x)
#define fil(x, y) fill(af(x), y)
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define ft front()
#define bk back()
#define f first
#define s second	
// function
#define fvoid function<void
#define fint function<int
#define fll function<ll
// rng
mt19937 g1;
void seed1(){g1.seed(time_t(0));}
void seed2(){g1.seed(chrono::steady_clock::now().time_since_epoch().count());}
ll randint(ll a,ll b){return uniform_int_distribution<ll>(a,b)(g1);}
// io
#define watch(x) cerr << (#x) << ": " << (x) << endl;

tcT> void rem_dup(vector<T> &v){sort(af(v));v.resize(unique(af(v))-v.begin());}
tcT> bool ckmin(T &a,const T &b){return b<a?a=b,1:0;}
tcT> bool ckmax(T &a,const T &b){return b>a?a=b,1:0;}
int pcnt(ll x){return __builtin_popcountll(x);}
int lg2(ll x){return 63 - __builtin_clzll(x);}
#define MOD1 1e9+7;
#define MOD2 998244353;
const int mod=MOD1;
ll fpow(ll a,ll b,ll m=mod){ll ret=1;while(b){if(b&1)ret=ret*a%m;a=a*a%m,b>>=1;}return ret;}
ll inv(ll x){return x?fpow(x,mod-2):1;}
const int MM = 1e5 + 5;

void solve(){
    int n, k;
    cin >> n >> k;
    vvi f(n + 1);
    for (int i = 1; i <= n * k; i++){
        int x;
        cin >> x;
        f[x].pb(i);
    }
    for (int i= 1; i <= n; i++){
        int id = 0;
        for (int j = 1; j < k - 1; j++){
            if (f[i][j + 1] - f[i][j] < f[i][id + 1] - f[i][id]) id = j;
        }
        cout << f[i][id] << ' ' << f[i][id + 1] << '\n';
    }
}
/*
 * edge case[1?, 2?]
 * do i rlly need to check everything???
 * i rlly need to check for long long sometimes.
 * guessforces
 */	
int main(){
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	seed2();
	int tests = 1;
	// cin >> tests;
	for (int i = 1; i <= tests; i++){
		//cout << "Case #" << i << ": ";
		solve();
	}
}
