#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e18;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (long long i = 0; i < v.size(); ++i) {
    os << v[i];
    if (i != v.size() - 1) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "{";
  for (auto it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  os << "{";
  for (auto it : v) {
    os << it.first << " : " << it.second;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(";
  os << v.first << ", " << v.second << ")";
  return os;
}
const long long mod = 1e9 + 7;
long long fin(long long a, vector<pair<long long, long long>>& par) {
  if (par[a].first == a)
    return a;
  else
    return fin(par[a].first, par);
}
bool uni(long long a, long long b, vector<pair<long long, long long>>& par) {
  long long o = fin(a, par);
  long long t = fin(b, par);
  if (o == t) return true;
  if (par[t].second > par[o].second) swap(o, t);
  if (par[t].second == par[o].second) par[o].second++;
  par[t].first = o;
  return false;
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long fin_par(vector<vector<long long>>& bin, long long a, long long p) {
  long long c = a;
  for (long long i = 0; i < 22; i++) {
    if ((1 << i) & p) {
      c = bin[c][i];
    }
  }
  return c;
}
long long lca(vector<vector<long long>>& bin, long long a, long long b,
              vector<long long>& d) {
  if (d[a] > d[b]) {
    swap(a, b);
  }
  long long dif = d[b] - d[a];
  long long prod = (1ll << 29);
  for (long long i = 29; i >= 0; i--) {
    if (dif & prod) {
      b = bin[b][i];
    }
    prod /= 2;
  }
  if (b == a) return a;
  prod = (1ll << 29);
  for (long long i = 29; i >= 0; i--) {
    if (bin[b][i] != bin[a][i]) {
      b = bin[b][i];
      a = bin[a][i];
    }
    prod /= 2;
  }
  return bin[a][0];
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long a[1000000];
map<long long, long long> mp;
set<long long> del;
vector<pair<long long, pair<long long, long long>>> segs;
set<long long> st;
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n * k; i++) {
    cin >> a[i];
  };
  long long maxi = (n + k - 2) / (k - 1);
  st.clear();
  mp.clear();
  del.clear();
  segs.clear();
  st.insert(0);
  long long pt = 0;
  for (long long i = 1; i < n + 1; i++) mp[i] = -1;
  for (long long i = 0; i < n * k; i++) {
    if (!del.count(a[i]) && mp[a[i]] != -1) {
      del.insert(a[i]);
      st.insert(i);
      segs.push_back({a[i], {mp[a[i]], i}});
      mp[a[i]] = -1;
    } else if (!del.count(a[i])) {
      mp[a[i]] = i;
    }
    while (st.size() > maxi) {
      st.erase(st.begin());
      while (pt < *st.begin()) {
        mp[a[pt]] = -1;
        pt++;
      }
    }
    if (segs.size() == n) break;
  }
  sort((segs).begin(), (segs).end());
  for (auto it : segs) {
    cout << it.second.first + 1 << " " << it.second.second + 1 << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  { solve(); }
}

### Prompt

Create a solution in cpp for 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 long long inf = 1e18;
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (long long i = 0; i < v.size(); ++i) {
    os << v[i];
    if (i != v.size() - 1) os << ", ";
  }
  os << "]";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "{";
  for (auto it : v) {
    os << it;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const map<T, S>& v) {
  os << "{";
  for (auto it : v) {
    os << it.first << " : " << it.second;
    if (it != *v.rbegin()) os << ", ";
  }
  os << "}";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << "(";
  os << v.first << ", " << v.second << ")";
  return os;
}
const long long mod = 1e9 + 7;
long long fin(long long a, vector<pair<long long, long long>>& par) {
  if (par[a].first == a)
    return a;
  else
    return fin(par[a].first, par);
}
bool uni(long long a, long long b, vector<pair<long long, long long>>& par) {
  long long o = fin(a, par);
  long long t = fin(b, par);
  if (o == t) return true;
  if (par[t].second > par[o].second) swap(o, t);
  if (par[t].second == par[o].second) par[o].second++;
  par[t].first = o;
  return false;
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long fin_par(vector<vector<long long>>& bin, long long a, long long p) {
  long long c = a;
  for (long long i = 0; i < 22; i++) {
    if ((1 << i) & p) {
      c = bin[c][i];
    }
  }
  return c;
}
long long lca(vector<vector<long long>>& bin, long long a, long long b,
              vector<long long>& d) {
  if (d[a] > d[b]) {
    swap(a, b);
  }
  long long dif = d[b] - d[a];
  long long prod = (1ll << 29);
  for (long long i = 29; i >= 0; i--) {
    if (dif & prod) {
      b = bin[b][i];
    }
    prod /= 2;
  }
  if (b == a) return a;
  prod = (1ll << 29);
  for (long long i = 29; i >= 0; i--) {
    if (bin[b][i] != bin[a][i]) {
      b = bin[b][i];
      a = bin[a][i];
    }
    prod /= 2;
  }
  return bin[a][0];
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long a[1000000];
map<long long, long long> mp;
set<long long> del;
vector<pair<long long, pair<long long, long long>>> segs;
set<long long> st;
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n * k; i++) {
    cin >> a[i];
  };
  long long maxi = (n + k - 2) / (k - 1);
  st.clear();
  mp.clear();
  del.clear();
  segs.clear();
  st.insert(0);
  long long pt = 0;
  for (long long i = 1; i < n + 1; i++) mp[i] = -1;
  for (long long i = 0; i < n * k; i++) {
    if (!del.count(a[i]) && mp[a[i]] != -1) {
      del.insert(a[i]);
      st.insert(i);
      segs.push_back({a[i], {mp[a[i]], i}});
      mp[a[i]] = -1;
    } else if (!del.count(a[i])) {
      mp[a[i]] = i;
    }
    while (st.size() > maxi) {
      st.erase(st.begin());
      while (pt < *st.begin()) {
        mp[a[pt]] = -1;
        pt++;
      }
    }
    if (segs.size() == n) break;
  }
  sort((segs).begin(), (segs).end());
  for (auto it : segs) {
    cout << it.second.first + 1 << " " << it.second.second + 1 << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  { solve(); }
}

```