#include <bits/stdc++.h>
using namespace std;
class SegTree {
  int leftMost, rightMost, toProp, mxVal;
  SegTree *lChild, *rChild;

 public:
  SegTree(int left, int right)
      : leftMost(left), rightMost(right), toProp(0), mxVal(0) {
    if (left != right) {
      int mid = (left + right) / 2;
      lChild = new SegTree(left, mid);
      rChild = new SegTree(mid + 1, right);
    } else {
      lChild = rChild = NULL;
    }
  }
  int getMax() { return mxVal + toProp; }
  void update(int left, int right, int delta = 1) {
    if (left > rightMost || right < leftMost)
      return;
    else if (left <= leftMost && right >= rightMost) {
      toProp += delta;
      return;
    } else {
      prop();
      lChild->update(left, right, delta);
      rChild->update(left, right, delta);
      recalc();
    }
  }
  int query(int left, int right) {
    if (left > rightMost || right < leftMost)
      return -1;
    else if (left <= leftMost && right >= rightMost) {
      return getMax();
    } else {
      prop();
      return max(lChild->query(left, right), rChild->query(left, right));
    }
  }
  void prop() {
    if (lChild) {
      lChild->toProp += toProp;
      rChild->toProp += toProp;
      toProp = 0;
    }
    recalc();
  }
  void recalc() {
    if (lChild) mxVal = max(lChild->getMax(), rChild->getMax());
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> inp(n * k);
  for (auto& u : inp) {
    cin >> u;
    --u;
  }
  SegTree st(0, n * k - 1);
  vector<int> lastSeen(n, -1);
  vector<bool> done(n, false);
  vector<pair<int, int>> ans(n);
  int maxAllowed = (n + k - 2) / (k - 1);
  for (int i = 0; i < n * k; i++) {
    if (lastSeen[inp[i]] == -1 || done[inp[i]]) {
      lastSeen[inp[i]] = i;
    } else {
      int chk = lastSeen[inp[i]];
      int getMost = st.query(chk, i);
      if (getMost < maxAllowed) {
        st.update(chk, i);
        done[inp[i]] = true;
        ans[inp[i]] = {chk + 1, i + 1};
      } else {
        lastSeen[inp[i]] = i;
      }
    }
  }
  for (auto u : ans) cout << u.first << " " << u.second << '\n';
  return 0;
}

### Prompt

Please formulate 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;
class SegTree {
  int leftMost, rightMost, toProp, mxVal;
  SegTree *lChild, *rChild;

 public:
  SegTree(int left, int right)
      : leftMost(left), rightMost(right), toProp(0), mxVal(0) {
    if (left != right) {
      int mid = (left + right) / 2;
      lChild = new SegTree(left, mid);
      rChild = new SegTree(mid + 1, right);
    } else {
      lChild = rChild = NULL;
    }
  }
  int getMax() { return mxVal + toProp; }
  void update(int left, int right, int delta = 1) {
    if (left > rightMost || right < leftMost)
      return;
    else if (left <= leftMost && right >= rightMost) {
      toProp += delta;
      return;
    } else {
      prop();
      lChild->update(left, right, delta);
      rChild->update(left, right, delta);
      recalc();
    }
  }
  int query(int left, int right) {
    if (left > rightMost || right < leftMost)
      return -1;
    else if (left <= leftMost && right >= rightMost) {
      return getMax();
    } else {
      prop();
      return max(lChild->query(left, right), rChild->query(left, right));
    }
  }
  void prop() {
    if (lChild) {
      lChild->toProp += toProp;
      rChild->toProp += toProp;
      toProp = 0;
    }
    recalc();
  }
  void recalc() {
    if (lChild) mxVal = max(lChild->getMax(), rChild->getMax());
  }
};
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> inp(n * k);
  for (auto& u : inp) {
    cin >> u;
    --u;
  }
  SegTree st(0, n * k - 1);
  vector<int> lastSeen(n, -1);
  vector<bool> done(n, false);
  vector<pair<int, int>> ans(n);
  int maxAllowed = (n + k - 2) / (k - 1);
  for (int i = 0; i < n * k; i++) {
    if (lastSeen[inp[i]] == -1 || done[inp[i]]) {
      lastSeen[inp[i]] = i;
    } else {
      int chk = lastSeen[inp[i]];
      int getMost = st.query(chk, i);
      if (getMost < maxAllowed) {
        st.update(chk, i);
        done[inp[i]] = true;
        ans[inp[i]] = {chk + 1, i + 1};
      } else {
        lastSeen[inp[i]] = i;
      }
    }
  }
  for (auto u : ans) cout << u.first << " " << u.second << '\n';
  return 0;
}

```