#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
}
template <typename T>
void print(T x) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
}
template <typename T>
void print(T x, char c) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
  putc(c);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MAXN = 100;
int n, k, c[MAXN * MAXN + 5], p[MAXN + 5], nxt[MAXN * MAXN + 5];
pair<int, int> res[MAXN + 5];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n * k; i++) scanf("%d", &c[i]);
  for (int L = 1; L <= n; L += k - 1) {
    int R = min(L + k - 2, n);
    vector<int> pos;
    for (int i = 1; i <= n * k; i++)
      if (L <= c[i] && c[i] <= R) pos.push_back(i);
    memset(p, 0, sizeof(p));
    memset(nxt, 0, sizeof(nxt));
    for (int i = 0; i < pos.size(); i++) {
      if (p[c[pos[i]]]) nxt[p[c[pos[i]]]] = pos[i];
      p[c[pos[i]]] = pos[i];
    }
    for (int i = L; i <= R; i++) nxt[p[i]] = n * k + 1;
    nxt[n * k + 1] = n * k + 1;
    memset(p, 0, sizeof(p));
    for (int i = 0; i < pos.size(); i++) {
      if (p[c[pos[i]]] && p[c[pos[i]]] != pos[i]) {
        res[c[pos[i]]] = make_pair(p[c[pos[i]]], pos[i]);
        p[c[pos[i]]] = 0;
        for (int j = L; j <= R; j++)
          if (p[j]) {
            while (p[j] <= pos[i]) p[j] = nxt[p[j]];
          }
      } else {
        if (!res[c[pos[i]]].first) p[c[pos[i]]] = pos[i];
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%d %d\n", res[i].first, res[i].second);
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T1, typename T2>
void chkmin(T1 &x, T2 y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
void chkmax(T1 &x, T2 y) {
  if (x < y) x = y;
}
namespace fastio {
char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
inline char getc() {
  return p1 == p2 &&
                 (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
             ? -1
             : *p1++;
}
inline void putc(char x) { (*p3++ = x); }
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T neg = 0;
  while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (neg) x = (~x) + 1;
}
template <typename T>
void recursive_print(T x) {
  return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
}
template <typename T>
void print(T x) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
}
template <typename T>
void print(T x, char c) {
  (!x) && (putc('0'), 0);
  (x < 0) && (putc('-'), x = ~x + 1);
  recursive_print(x);
  putc(c);
}
void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
}  // namespace fastio
const int MAXN = 100;
int n, k, c[MAXN * MAXN + 5], p[MAXN + 5], nxt[MAXN * MAXN + 5];
pair<int, int> res[MAXN + 5];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n * k; i++) scanf("%d", &c[i]);
  for (int L = 1; L <= n; L += k - 1) {
    int R = min(L + k - 2, n);
    vector<int> pos;
    for (int i = 1; i <= n * k; i++)
      if (L <= c[i] && c[i] <= R) pos.push_back(i);
    memset(p, 0, sizeof(p));
    memset(nxt, 0, sizeof(nxt));
    for (int i = 0; i < pos.size(); i++) {
      if (p[c[pos[i]]]) nxt[p[c[pos[i]]]] = pos[i];
      p[c[pos[i]]] = pos[i];
    }
    for (int i = L; i <= R; i++) nxt[p[i]] = n * k + 1;
    nxt[n * k + 1] = n * k + 1;
    memset(p, 0, sizeof(p));
    for (int i = 0; i < pos.size(); i++) {
      if (p[c[pos[i]]] && p[c[pos[i]]] != pos[i]) {
        res[c[pos[i]]] = make_pair(p[c[pos[i]]], pos[i]);
        p[c[pos[i]]] = 0;
        for (int j = L; j <= R; j++)
          if (p[j]) {
            while (p[j] <= pos[i]) p[j] = nxt[p[j]];
          }
      } else {
        if (!res[c[pos[i]]].first) p[c[pos[i]]] = pos[i];
      }
    }
  }
  for (int i = 1; i <= n; i++) printf("%d %d\n", res[i].first, res[i].second);
  return 0;
}

```