#include <bits/stdc++.h>
using namespace std;
inline int getint() {
  int n;
  scanf("%d", &n);
  return n;
}
vector<int> a[105];
vector<pair<int, int> > b;
int sg[100005], add[100005];
bool lazy[100005];
void push_down(int rt) {
  lazy[rt << 1] = lazy[rt << 1 | 1] = true;
  add[rt << 1] += add[rt];
  add[rt << 1 | 1] += add[rt];
  sg[rt << 1] += add[rt];
  sg[rt << 1 | 1] += add[rt];
  add[rt] = 0;
  lazy[rt] = false;
}
void push_up(int rt) { sg[rt] = max(sg[rt << 1], sg[rt << 1 | 1]); }
int get_max(int L, int R, int rt, int l, int r) {
  if (L <= l && r <= R) return sg[rt];
  if (lazy[rt]) push_down(rt);
  int ans = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) ans = max(ans, get_max(L, R, rt << 1, l, mid));
  if (R > mid) ans = max(ans, get_max(L, R, rt << 1 | 1, mid + 1, r));
  return ans;
}
void add_one(int L, int R, int rt, int l, int r) {
  if (L <= l && r <= R) {
    lazy[rt] = true;
    add[rt] += 1;
    sg[rt] += 1;
    return;
  }
  if (lazy[rt]) push_down(rt);
  int mid = (l + r) >> 1;
  if (L <= mid) add_one(L, R, rt << 1, l, mid);
  if (R > mid) add_one(L, R, rt << 1 | 1, mid + 1, r);
  push_up(rt);
}
bool cmp(const pair<int, int> &x, const pair<int, int> &y) {
  return x.second < y.second;
}
bool used[105];
int record[10005];
vector<pair<int, int> > ans;
int main() {
  int n = getint(), K = getint();
  for (int i = 1; i <= n * K; i++) {
    int c = getint();
    record[i] = c;
    a[c].emplace_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < a[i].size(); j++) {
      for (int k = j + 1; k < a[i].size(); k++) {
        b.emplace_back(pair<int, int>(a[i][j], a[i][k]));
      }
    }
  }
  sort(b.begin(), b.end(), cmp);
  int M = int(n + K - 2) / (K - 1);
  int cnt = n;
  ans.resize(n + 1);
  for (int i = 0; i < b.size(); i++) {
    if (cnt == 0) break;
    int col = record[b[i].first];
    if (used[col]) continue;
    if (get_max(b[i].first, b[i].second, 1, 1, n * K) < M) {
      ans[col] = pair<int, int>(b[i].first, b[i].second);
      cnt--;
      used[col] = true;
      add_one(b[i].first, b[i].second, 1, 1, n * K);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", ans[i].first, ans[i].second);
  }
  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>
using namespace std;
inline int getint() {
  int n;
  scanf("%d", &n);
  return n;
}
vector<int> a[105];
vector<pair<int, int> > b;
int sg[100005], add[100005];
bool lazy[100005];
void push_down(int rt) {
  lazy[rt << 1] = lazy[rt << 1 | 1] = true;
  add[rt << 1] += add[rt];
  add[rt << 1 | 1] += add[rt];
  sg[rt << 1] += add[rt];
  sg[rt << 1 | 1] += add[rt];
  add[rt] = 0;
  lazy[rt] = false;
}
void push_up(int rt) { sg[rt] = max(sg[rt << 1], sg[rt << 1 | 1]); }
int get_max(int L, int R, int rt, int l, int r) {
  if (L <= l && r <= R) return sg[rt];
  if (lazy[rt]) push_down(rt);
  int ans = 0;
  int mid = (l + r) >> 1;
  if (L <= mid) ans = max(ans, get_max(L, R, rt << 1, l, mid));
  if (R > mid) ans = max(ans, get_max(L, R, rt << 1 | 1, mid + 1, r));
  return ans;
}
void add_one(int L, int R, int rt, int l, int r) {
  if (L <= l && r <= R) {
    lazy[rt] = true;
    add[rt] += 1;
    sg[rt] += 1;
    return;
  }
  if (lazy[rt]) push_down(rt);
  int mid = (l + r) >> 1;
  if (L <= mid) add_one(L, R, rt << 1, l, mid);
  if (R > mid) add_one(L, R, rt << 1 | 1, mid + 1, r);
  push_up(rt);
}
bool cmp(const pair<int, int> &x, const pair<int, int> &y) {
  return x.second < y.second;
}
bool used[105];
int record[10005];
vector<pair<int, int> > ans;
int main() {
  int n = getint(), K = getint();
  for (int i = 1; i <= n * K; i++) {
    int c = getint();
    record[i] = c;
    a[c].emplace_back(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < a[i].size(); j++) {
      for (int k = j + 1; k < a[i].size(); k++) {
        b.emplace_back(pair<int, int>(a[i][j], a[i][k]));
      }
    }
  }
  sort(b.begin(), b.end(), cmp);
  int M = int(n + K - 2) / (K - 1);
  int cnt = n;
  ans.resize(n + 1);
  for (int i = 0; i < b.size(); i++) {
    if (cnt == 0) break;
    int col = record[b[i].first];
    if (used[col]) continue;
    if (get_max(b[i].first, b[i].second, 1, 1, n * K) < M) {
      ans[col] = pair<int, int>(b[i].first, b[i].second);
      cnt--;
      used[col] = true;
      add_one(b[i].first, b[i].second, 1, 1, n * K);
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d %d\n", ans[i].first, ans[i].second);
  }
  return 0;
}

```