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>
using namespace std;
#define fo(i,n) for(i=0;i<n;i++)
#define fo2(i,start,end) for(i=start;i<=end;i++)
#define rfo(i,n) for(i=n-1;i>=0;i--)
#define ll long long int
#define deb(x) cout<<#x<<"="<<x<<endl
#define deb2(x,y) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define sortall(x) sort(all(x))
#define tr(it,a) for(auto it=a.begin(); it!=a.end();it++)
#define nl '\n'
#define PI 3.1415926535897932384626
#define mod 1000000007
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pl>		vpl;
const int N=1e6+10;

void solve(){
    int i,j;
    int n,k,m,val;
    cin>>n>>k;
    m=n*k;
    val=(n+k-2)/(k-1);
    // deb(val);
    int a[m+1],b[m+1]={0};
    fo2(i,1,m){
        cin>>a[i];
    }
    map<int,int> mapp;
    set<int> se;
    vpii v;
    fo2(i,1,m){
        // deb2(a[i],mapp[a[i]]);
        if(se.find(a[i])!=se.end()){
            continue;
        }
        if(mapp[a[i]]==0){
            mapp[a[i]]=i;
            continue;
        }
        for(j=mapp[a[i]];j<=i;j++){
            if(b[j]>=val){
                break;
            }
        }
        // deb2(j,i);
        if(j!=i+1){
            mapp[a[i]]=i;
            continue;
        }
        v.pb({mapp[a[i]],i});
        se.insert({a[i]});
        for(j=mapp[a[i]];j<=i;j++){
            b[j]++;
        }
    }
    fo(i,v.size()){
        cout<<v[i].F<<" "<<v[i].S<<nl;
    }   
}

int main(){
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t=1;
    // cin>>t;
    while(t--){
        solve();
    }
    
}

