#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long int;
using ld = long double;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
struct edge {
  ll to;
  ll cost;
};
using graph = vector<vector<edge>>;
const ll INF = 1e18;
inline void chmax(ll& a, ll b) { a = max(a, b); }
inline void chmin(ll& a, ll b) { a = min(a, b); }
long double pi = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll dx[4]{1, 0, -1, 0};
ll dy[4]{0, 1, 0, -1};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n, k;
  cin >> n >> k;
  ll m = n * k;
  vector<ll> A(m);
  for (ll i = 0; i < (m); ++i) cin >> A[i];
  vector<vector<ll>> B(n);
  for (ll i = 0; i < (m); ++i) B[A[i] - 1].push_back(i);
  vector<vector<P>> unti(n);
  for (ll i = 0; i < (n); ++i)
    for (ll j = 0; j < (k - 1); ++j) {
      ll x = B[i][j];
      ll y = B[i][j + 1];
      unti.push_back({make_pair(x, y)});
    }
  vector<ll> seen(n);
  vector<P> ans(n);
  ll cnt = 0;
  bool ok = 0;
  for (ll i = 0; i < (n); ++i) {
    ll now = -1;
    while (1) {
      ll ma = INF;
      ll next = -1;
      ll soeji = -1;
      for (ll j = 0; j < (n); ++j) {
        if (seen[j]) continue;
        ll y = upper_bound(B[j].begin(), B[j].end(), now) - B[j].begin();
        if (y == k - 1 || y == k) continue;
        ll x = B[j][y + 1];
        if (x < ma) {
          ma = x;
          next = j;
          soeji = y;
        }
      }
      if (ma == INF) break;
      seen[next] = true;
      ans[next].first = B[next][soeji];
      ans[next].second = B[next][soeji + 1];
      now = ma;
      cnt++;
      if (cnt == n) {
        ok = 1;
        break;
      }
    }
    if (ok) break;
  }
  for (ll i = 0; i < (n); ++i)
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = long long int;
using ld = long double;
using P = pair<ll, ll>;
using Graph = vector<vector<ll>>;
struct edge {
  ll to;
  ll cost;
};
using graph = vector<vector<edge>>;
const ll INF = 1e18;
inline void chmax(ll& a, ll b) { a = max(a, b); }
inline void chmin(ll& a, ll b) { a = min(a, b); }
long double pi = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll dx[4]{1, 0, -1, 0};
ll dy[4]{0, 1, 0, -1};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n, k;
  cin >> n >> k;
  ll m = n * k;
  vector<ll> A(m);
  for (ll i = 0; i < (m); ++i) cin >> A[i];
  vector<vector<ll>> B(n);
  for (ll i = 0; i < (m); ++i) B[A[i] - 1].push_back(i);
  vector<vector<P>> unti(n);
  for (ll i = 0; i < (n); ++i)
    for (ll j = 0; j < (k - 1); ++j) {
      ll x = B[i][j];
      ll y = B[i][j + 1];
      unti.push_back({make_pair(x, y)});
    }
  vector<ll> seen(n);
  vector<P> ans(n);
  ll cnt = 0;
  bool ok = 0;
  for (ll i = 0; i < (n); ++i) {
    ll now = -1;
    while (1) {
      ll ma = INF;
      ll next = -1;
      ll soeji = -1;
      for (ll j = 0; j < (n); ++j) {
        if (seen[j]) continue;
        ll y = upper_bound(B[j].begin(), B[j].end(), now) - B[j].begin();
        if (y == k - 1 || y == k) continue;
        ll x = B[j][y + 1];
        if (x < ma) {
          ma = x;
          next = j;
          soeji = y;
        }
      }
      if (ma == INF) break;
      seen[next] = true;
      ans[next].first = B[next][soeji];
      ans[next].second = B[next][soeji + 1];
      now = ma;
      cnt++;
      if (cnt == n) {
        ok = 1;
        break;
      }
    }
    if (ok) break;
  }
  for (ll i = 0; i < (n); ++i)
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << endl;
  return 0;
}

```