#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
using namespace std;
long long mod = 998244353;
int mul(long long a, long long b) { return (a * b) % mod; }
int binexp(long long a, long long b) {
  if (b == 0) return 1;
  long long half = binexp(a, b / 2);
  if ((b % 2) == 0)
    return mul(half, half);
  else
    return mul(a, mul(half, half));
}
void prop(int node, int tree[], int delta[]) {
  delta[node << 1] += delta[node];
  tree[node << 1] += delta[node];
  delta[node << 1 | 1] += delta[node];
  tree[node << 1 | 1] += delta[node];
  delta[node] = 0;
}
void update(int node, int l, int r, int start, int end, int tree[],
            int delta[]) {
  if (l > end || r < start || start > end) return;
  if (l <= start && end <= r) {
    tree[node] += 1;
    delta[node] += 1;
    return;
  }
  prop(node, tree, delta);
  int mid = start + (end - start) / 2;
  update(node << 1, l, r, start, mid, tree, delta);
  update(node << 1 | 1, l, r, mid + 1, end, tree, delta);
  tree[node] = max(tree[node << 1], tree[node << 1 | 1]);
}
int query(int node, int l, int r, int start, int end, int tree[], int delta[]) {
  if (l > end || r < start || start > end) return INT_MIN;
  if (l <= start && end <= r) return tree[node];
  prop(node, tree, delta);
  int mid = start + (end - start) / 2;
  int p1 = query(node << 1, l, r, start, mid, tree, delta);
  int p2 = query(node << 1 | 1, l, r, mid + 1, end, tree, delta);
  tree[node] = max(tree[node << 1], tree[node << 1 | 1]);
  return max(p1, p2);
}
bool cmp(pair<int, pair<int, int>> p1, pair<int, pair<int, int>> p2) {
  return (p1.first < p2.first) ||
         ((p1.first == p2.first && (p1.second.first > p2.second.first ||
                                    (p1.second.first == p2.second.first &&
                                     p1.second.second > p2.second.second))));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int arr[n * k];
    for (long long i = 0; i < n * k; ++i) {
      cin >> arr[i];
    }
    int tree[4 * n * k];
    memset(tree, 0, sizeof(tree));
    int last[n + 1];
    memset(last, -1, sizeof(last));
    int vis[n * k];
    memset(vis, false, sizeof(vis));
    int delta[4 * n * k];
    memset(delta, 0, sizeof(delta));
    vector<pair<int, pair<int, int>>> ans;
    for (long long i = 0; i < n * k; ++i) {
      if (last[arr[i]] == -1) {
        last[arr[i]] = i;
      } else if (vis[last[arr[i]]])
        continue;
      else {
        int size = (n + k - 2) / (k - 1);
        int temp = query(1, last[arr[i]], i, 0, n * k - 1, tree, delta);
        if (temp < size) {
          update(1, last[arr[i]], i, 0, n * k - 1, tree, delta);
          vis[i] = true;
          ans.push_back({arr[i], {last[arr[i]], i}});
        }
        last[arr[i]] = i;
      }
    }
    sort(ans.begin(), ans.end());
    for (long long i = 0; i < min(n, (int)ans.size()); ++i) {
      cout << ans[i].second.first + 1 << " " << ans[i].second.second + 1
           << endl;
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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("O3")
#pragma GCC target("avx2")
using namespace std;
long long mod = 998244353;
int mul(long long a, long long b) { return (a * b) % mod; }
int binexp(long long a, long long b) {
  if (b == 0) return 1;
  long long half = binexp(a, b / 2);
  if ((b % 2) == 0)
    return mul(half, half);
  else
    return mul(a, mul(half, half));
}
void prop(int node, int tree[], int delta[]) {
  delta[node << 1] += delta[node];
  tree[node << 1] += delta[node];
  delta[node << 1 | 1] += delta[node];
  tree[node << 1 | 1] += delta[node];
  delta[node] = 0;
}
void update(int node, int l, int r, int start, int end, int tree[],
            int delta[]) {
  if (l > end || r < start || start > end) return;
  if (l <= start && end <= r) {
    tree[node] += 1;
    delta[node] += 1;
    return;
  }
  prop(node, tree, delta);
  int mid = start + (end - start) / 2;
  update(node << 1, l, r, start, mid, tree, delta);
  update(node << 1 | 1, l, r, mid + 1, end, tree, delta);
  tree[node] = max(tree[node << 1], tree[node << 1 | 1]);
}
int query(int node, int l, int r, int start, int end, int tree[], int delta[]) {
  if (l > end || r < start || start > end) return INT_MIN;
  if (l <= start && end <= r) return tree[node];
  prop(node, tree, delta);
  int mid = start + (end - start) / 2;
  int p1 = query(node << 1, l, r, start, mid, tree, delta);
  int p2 = query(node << 1 | 1, l, r, mid + 1, end, tree, delta);
  tree[node] = max(tree[node << 1], tree[node << 1 | 1]);
  return max(p1, p2);
}
bool cmp(pair<int, pair<int, int>> p1, pair<int, pair<int, int>> p2) {
  return (p1.first < p2.first) ||
         ((p1.first == p2.first && (p1.second.first > p2.second.first ||
                                    (p1.second.first == p2.second.first &&
                                     p1.second.second > p2.second.second))));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int arr[n * k];
    for (long long i = 0; i < n * k; ++i) {
      cin >> arr[i];
    }
    int tree[4 * n * k];
    memset(tree, 0, sizeof(tree));
    int last[n + 1];
    memset(last, -1, sizeof(last));
    int vis[n * k];
    memset(vis, false, sizeof(vis));
    int delta[4 * n * k];
    memset(delta, 0, sizeof(delta));
    vector<pair<int, pair<int, int>>> ans;
    for (long long i = 0; i < n * k; ++i) {
      if (last[arr[i]] == -1) {
        last[arr[i]] = i;
      } else if (vis[last[arr[i]]])
        continue;
      else {
        int size = (n + k - 2) / (k - 1);
        int temp = query(1, last[arr[i]], i, 0, n * k - 1, tree, delta);
        if (temp < size) {
          update(1, last[arr[i]], i, 0, n * k - 1, tree, delta);
          vis[i] = true;
          ans.push_back({arr[i], {last[arr[i]], i}});
        }
        last[arr[i]] = i;
      }
    }
    sort(ans.begin(), ans.end());
    for (long long i = 0; i < min(n, (int)ans.size()); ++i) {
      cout << ans[i].second.first + 1 << " " << ans[i].second.second + 1
           << endl;
    }
  }
}

```