#include <bits/stdc++.h>
using namespace std;
array<long long int, 2> ans[103];
long long int prev_pos[103];
bool flg[103];
long long int st[80303];
long long int lazy[80303];
long long int a[10003];
void do_lazy(long long int node, long long int l, long long int r) {
  st[node] += lazy[node];
  if (l != r) {
    lazy[2ll * node + 1ll] += lazy[node];
    lazy[2ll * node + 2ll] += lazy[node];
  }
  lazy[node] = 0;
}
void upd(long long int node, long long int l, long long int r, long long int ql,
         long long int qr, long long int val) {
  do_lazy(node, l, r);
  if (l > r || l > qr || r < ql) {
    return;
  }
  if (l >= ql && r <= qr) {
    lazy[node] += val;
    do_lazy(node, l, r);
    return;
  }
  long long int mid = (l + r) / 2;
  upd(2ll * node + 1ll, l, mid, ql, qr, val);
  upd(2ll * node + 2ll, mid + 1ll, r, ql, qr, val);
  st[node] = max(st[2ll * node + 1ll], st[2ll * node + 2ll]);
}
long long int query(long long int node, long long int l, long long int r,
                    long long int ql, long long int qr) {
  do_lazy(node, l, r);
  if (l > r || l > qr || r < ql) {
    return 0;
  }
  if (l >= ql && r <= qr) {
    return st[node];
  }
  long long int mid = (l + r) / 2;
  return max(query(2ll * node + 1ll, l, mid, ql, qr),
             query(2ll * node + 2ll, mid + 1ll, r, ql, qr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  for (long long int i = 1ll; i <= n * k; i++) {
    cin >> a[i];
  }
  for (long long int i = 1ll; i <= n * k; i++) {
    if (flg[a[i]]) {
      continue;
    }
    if (prev_pos[a[i]] == 0) {
      prev_pos[a[i]] = i;
      continue;
    }
    long long int val = query(1, 1, n * k, prev_pos[a[i]], i - 1ll);
    if (val >= ((n + k - 2ll) / (k - 1ll))) {
      prev_pos[a[i]] = i;
    } else {
      upd(1, 1, n * k, prev_pos[a[i]], i, 1);
      flg[a[i]] = 1;
      ans[a[i]] = {prev_pos[a[i]], i};
    }
  }
  for (long long int i = 1ll; i <= n; i++) {
    cout << ans[i][0] << " " << ans[i][1] << "\n";
  }
}

### 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>
using namespace std;
array<long long int, 2> ans[103];
long long int prev_pos[103];
bool flg[103];
long long int st[80303];
long long int lazy[80303];
long long int a[10003];
void do_lazy(long long int node, long long int l, long long int r) {
  st[node] += lazy[node];
  if (l != r) {
    lazy[2ll * node + 1ll] += lazy[node];
    lazy[2ll * node + 2ll] += lazy[node];
  }
  lazy[node] = 0;
}
void upd(long long int node, long long int l, long long int r, long long int ql,
         long long int qr, long long int val) {
  do_lazy(node, l, r);
  if (l > r || l > qr || r < ql) {
    return;
  }
  if (l >= ql && r <= qr) {
    lazy[node] += val;
    do_lazy(node, l, r);
    return;
  }
  long long int mid = (l + r) / 2;
  upd(2ll * node + 1ll, l, mid, ql, qr, val);
  upd(2ll * node + 2ll, mid + 1ll, r, ql, qr, val);
  st[node] = max(st[2ll * node + 1ll], st[2ll * node + 2ll]);
}
long long int query(long long int node, long long int l, long long int r,
                    long long int ql, long long int qr) {
  do_lazy(node, l, r);
  if (l > r || l > qr || r < ql) {
    return 0;
  }
  if (l >= ql && r <= qr) {
    return st[node];
  }
  long long int mid = (l + r) / 2;
  return max(query(2ll * node + 1ll, l, mid, ql, qr),
             query(2ll * node + 2ll, mid + 1ll, r, ql, qr));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, k;
  cin >> n >> k;
  for (long long int i = 1ll; i <= n * k; i++) {
    cin >> a[i];
  }
  for (long long int i = 1ll; i <= n * k; i++) {
    if (flg[a[i]]) {
      continue;
    }
    if (prev_pos[a[i]] == 0) {
      prev_pos[a[i]] = i;
      continue;
    }
    long long int val = query(1, 1, n * k, prev_pos[a[i]], i - 1ll);
    if (val >= ((n + k - 2ll) / (k - 1ll))) {
      prev_pos[a[i]] = i;
    } else {
      upd(1, 1, n * k, prev_pos[a[i]], i, 1);
      flg[a[i]] = 1;
      ans[a[i]] = {prev_pos[a[i]], i};
    }
  }
  for (long long int i = 1ll; i <= n; i++) {
    cout << ans[i][0] << " " << ans[i][1] << "\n";
  }
}

```