#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 998244353;
const long long maxn = 1.2e6 + 107;
const double pi = acos(-1.0);
const double eps = 0.0000000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void pr2(T x, unsigned long long k = 64) {
  unsigned long long i;
  for (i = 0; i < k; i++) fprintf(stderr, "%d", (int)((x >> i) & 1));
  putchar(' ');
}
template <typename T>
inline void add_(T &A, int B, long long MOD = M) {
  A += B;
  (A >= MOD) && (A -= MOD);
}
template <typename T>
inline void mul_(T &A, long long B, long long MOD = M) {
  A = (A * B) % MOD;
}
template <typename T>
inline void mod_(T &A, long long MOD = M) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline void max_(T &A, T B) {
  (A < B) && (A = B);
}
template <typename T>
inline void min_(T &A, T B) {
  (A > B) && (A = B);
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
inline long long powMM(long long a, long long b, long long mod = M) {
  long long ret = 1;
  for (; b; b >>= 1ll, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int startTime;
void startTimer() { startTime = clock(); }
void printTimer() {
  fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
}
int A[maxn], R[maxn], id[maxn];
pair<int, int> ans[maxn];
set<int> Q;
int main() {
  int T, _;
  T = 1;
  for (_ = 1; _ <= T; _++) {
    int n, i, K;
    scanf("%d%d", &n, &K);
    int maxCount = (n + K - 1 - 1) / (K - 1);
    for (i = 1; i <= n * K; i++) scanf("%d", &A[i]);
    for (i = n * K; i >= 1; i--) {
      R[i] = id[A[i]];
      id[A[i]] = i;
    }
    for (i = 1; i <= n * K; i++) {
      if (ans[A[i]].second == i) Q.erase(i);
      if (ans[A[i]].first == 0) ans[A[i]] = make_pair(i, R[i]), Q.insert(R[i]);
      while ((int)Q.size() > maxCount) {
        int x = *Q.rbegin();
        Q.erase(x);
        ans[A[x]] = make_pair(0, 0);
      }
    }
    for (i = 1; i <= n; i++) printf("%d %d\n", ans[i].first, ans[i].second);
  }
}

### 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>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long INFF = 0x3f3f3f3f3f3f3f3fll;
const long long M = 998244353;
const long long maxn = 1.2e6 + 107;
const double pi = acos(-1.0);
const double eps = 0.0000000001;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
template <typename T>
inline void pr2(T x, unsigned long long k = 64) {
  unsigned long long i;
  for (i = 0; i < k; i++) fprintf(stderr, "%d", (int)((x >> i) & 1));
  putchar(' ');
}
template <typename T>
inline void add_(T &A, int B, long long MOD = M) {
  A += B;
  (A >= MOD) && (A -= MOD);
}
template <typename T>
inline void mul_(T &A, long long B, long long MOD = M) {
  A = (A * B) % MOD;
}
template <typename T>
inline void mod_(T &A, long long MOD = M) {
  A %= MOD;
  A += MOD;
  A %= MOD;
}
template <typename T>
inline void max_(T &A, T B) {
  (A < B) && (A = B);
}
template <typename T>
inline void min_(T &A, T B) {
  (A > B) && (A = B);
}
template <typename T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
inline long long powMM(long long a, long long b, long long mod = M) {
  long long ret = 1;
  for (; b; b >>= 1ll, a = a * a % mod)
    if (b & 1) ret = ret * a % mod;
  return ret;
}
int startTime;
void startTimer() { startTime = clock(); }
void printTimer() {
  fprintf(stderr, "/--- Time: %ld milliseconds ---/\n", clock() - startTime);
}
int A[maxn], R[maxn], id[maxn];
pair<int, int> ans[maxn];
set<int> Q;
int main() {
  int T, _;
  T = 1;
  for (_ = 1; _ <= T; _++) {
    int n, i, K;
    scanf("%d%d", &n, &K);
    int maxCount = (n + K - 1 - 1) / (K - 1);
    for (i = 1; i <= n * K; i++) scanf("%d", &A[i]);
    for (i = n * K; i >= 1; i--) {
      R[i] = id[A[i]];
      id[A[i]] = i;
    }
    for (i = 1; i <= n * K; i++) {
      if (ans[A[i]].second == i) Q.erase(i);
      if (ans[A[i]].first == 0) ans[A[i]] = make_pair(i, R[i]), Q.insert(R[i]);
      while ((int)Q.size() > maxCount) {
        int x = *Q.rbegin();
        Q.erase(x);
        ans[A[x]] = make_pair(0, 0);
      }
    }
    for (i = 1; i <= n; i++) printf("%d %d\n", ans[i].first, ans[i].second);
  }
}

```