#include <bits/stdc++.h>
using namespace std;
const int maxn = 105, pot = 16384;
struct tournament {
  int t[pot * 2];
  int prop[pot * 2];
  void propagate(int x) {
    if (prop[x]) {
      t[x] += prop[x];
      if (x < pot) {
        prop[x * 2] += prop[x];
        prop[x * 2 + 1] += prop[x];
      }
      prop[x] = 0;
    }
  }
  void update(int x, int val) {
    t[x] += val;
    for (x /= 2; x > 0; x /= 2) {
      propagate(x * 2);
      propagate(x * 2 + 1);
      t[x] = max(t[x * 2], t[x * 2 + 1]);
    }
  }
  void update2(int L, int D, int x, int l, int d, int val) {
    propagate(x);
    if (L >= l && D <= d) {
      update(x, val);
      if (x < pot) {
        prop[x * 2] += val;
        prop[x * 2 + 1] += val;
      }
      return;
    }
    if ((L + D) / 2 >= l) {
      update2(L, (L + D) / 2, x * 2, l, d, val);
    }
    if ((L + D) / 2 + 1 <= d) {
      update2((L + D) / 2 + 1, D, x * 2 + 1, l, d, val);
    }
  }
  int query(int L, int D, int x, int l, int d) {
    propagate(x);
    if (L >= l && D <= d) {
      return t[x];
    }
    int lijeva = 0, desna = 0;
    if ((L + D) / 2 >= l) {
      lijeva = query(L, (L + D) / 2, x * 2, l, d);
    }
    if ((L + D) / 2 + 1 <= d) {
      desna = query((L + D) / 2 + 1, D, x * 2 + 1, l, d);
    }
    return max(lijeva, desna);
  }
};
tournament T;
vector<pair<int, int> > v;
int a[maxn * maxn];
int pos[maxn];
int ind[maxn];
vector<int> boja[maxn];
bool cmp(int x, int y) { return boja[x][ind[x]] > boja[y][ind[y]]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n * k; i++) {
    cin >> a[i];
    a[i]--;
    boja[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    pos[i] = i;
    ind[i] = 1;
    T.update2(0, pot - 1, 1, boja[i][0], boja[i][1], 1);
  }
  bool p = 1;
  int maksi = (n + k - 2) / (k - 1);
  int x;
  while (p) {
    p = 0;
    sort(pos, pos + n, cmp);
    for (int i = 0; i < n; i++) {
      x = pos[i];
      if (T.query(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]]) >
          maksi) {
        p = 1;
        T.update2(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]], -1);
        ind[x]++;
        T.update2(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]], 1);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << boja[i][ind[i] - 1] + 1 << ' ' << boja[i][ind[i]] + 1 << '\n';
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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;
const int maxn = 105, pot = 16384;
struct tournament {
  int t[pot * 2];
  int prop[pot * 2];
  void propagate(int x) {
    if (prop[x]) {
      t[x] += prop[x];
      if (x < pot) {
        prop[x * 2] += prop[x];
        prop[x * 2 + 1] += prop[x];
      }
      prop[x] = 0;
    }
  }
  void update(int x, int val) {
    t[x] += val;
    for (x /= 2; x > 0; x /= 2) {
      propagate(x * 2);
      propagate(x * 2 + 1);
      t[x] = max(t[x * 2], t[x * 2 + 1]);
    }
  }
  void update2(int L, int D, int x, int l, int d, int val) {
    propagate(x);
    if (L >= l && D <= d) {
      update(x, val);
      if (x < pot) {
        prop[x * 2] += val;
        prop[x * 2 + 1] += val;
      }
      return;
    }
    if ((L + D) / 2 >= l) {
      update2(L, (L + D) / 2, x * 2, l, d, val);
    }
    if ((L + D) / 2 + 1 <= d) {
      update2((L + D) / 2 + 1, D, x * 2 + 1, l, d, val);
    }
  }
  int query(int L, int D, int x, int l, int d) {
    propagate(x);
    if (L >= l && D <= d) {
      return t[x];
    }
    int lijeva = 0, desna = 0;
    if ((L + D) / 2 >= l) {
      lijeva = query(L, (L + D) / 2, x * 2, l, d);
    }
    if ((L + D) / 2 + 1 <= d) {
      desna = query((L + D) / 2 + 1, D, x * 2 + 1, l, d);
    }
    return max(lijeva, desna);
  }
};
tournament T;
vector<pair<int, int> > v;
int a[maxn * maxn];
int pos[maxn];
int ind[maxn];
vector<int> boja[maxn];
bool cmp(int x, int y) { return boja[x][ind[x]] > boja[y][ind[y]]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n * k; i++) {
    cin >> a[i];
    a[i]--;
    boja[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    pos[i] = i;
    ind[i] = 1;
    T.update2(0, pot - 1, 1, boja[i][0], boja[i][1], 1);
  }
  bool p = 1;
  int maksi = (n + k - 2) / (k - 1);
  int x;
  while (p) {
    p = 0;
    sort(pos, pos + n, cmp);
    for (int i = 0; i < n; i++) {
      x = pos[i];
      if (T.query(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]]) >
          maksi) {
        p = 1;
        T.update2(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]], -1);
        ind[x]++;
        T.update2(0, pot - 1, 1, boja[x][ind[x] - 1], boja[x][ind[x]], 1);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    cout << boja[i][ind[i] - 1] + 1 << ' ' << boja[i][ind[i]] + 1 << '\n';
  }
  return 0;
}

```