description: 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>
solution: #include <bits/stdc++.h>
  using namespace std;
  using uint = unsigned int;
  using ll = long long;
  using ull = unsigned long long;
  using db = double;
  using ld = long double;
  using pii = pair<int, int>;
  using pll = pair<ll, ll>;
  using pdd = pair<double, double>;
  using vint = vector<int>;
  using vpii = vector<pii>;
  pii operator+(pii a, pii b) { return {a.first + b.first, a.second + b.second}; }
  pll operator+(pll a, pll b) { return {a.first + b.first, a.second + b.second}; }
  namespace IO {
  char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 24], *O = obuf;
  inline ll read() {
    ll x = 0;
    bool sign = 0;
    char s =
        (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
             ? EOF
             : *p1++);
    while (!isdigit(s))
      sign |= s == '-',
          s = (p1 == p2 &&
                       (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
                   ? EOF
                   : *p1++);
    while (isdigit(s))
      x = (x << 1) + (x << 3) + (s - '0'),
      s = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 22, stdin), p1 == p2)
               ? EOF
               : *p1++);
    return sign ? -x : x;
  }
  inline void print(ll x) {
    if (x < 0)
      (*O++ = '-'), print(-x);
    else {
      if (x > 9) print(x / 10);
      (*O++ = x % 10 + '0');
    }
  }
  }  // namespace IO
  const int mod = 998244353;
  const long double Pi = acos(-1);
  namespace math {
  const int MAXN = 2e6 + 5;
  ll ksm(ll a, ll b, ll p) {
    ll s = 1, m = a;
    while (b) {
      if (b & 1) s = s * m % p;
      m = m * m % p, b >>= 1;
    }
    return s;
  }
  ll ksm(ll a, ll b) {
    ll s = 1, m = a;
    while (b) {
      if (b & 1) s = s * m % mod;
      m = m * m % mod, b >>= 1;
    }
    return s;
  }
  ll inv(ll x) { return ksm(x, mod - 2); }
  ll fc[MAXN], ifc[MAXN];
  void init(int n) {
    assert(n < MAXN);
    fc[0] = 1;
    for (int i = 1; i <= n; i++) fc[i] = fc[i - 1] * i % mod;
    ifc[n] = inv(fc[n]);
    for (int i = n - 1; i >= 0; i--) ifc[i] = ifc[i + 1] * (i + 1) % mod;
  }
  ll bin(ll n, ll m) { return fc[n] * ifc[m] % mod * ifc[n - m] % mod; }
  }  // namespace math
  mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
  using namespace IO;
  const int N = 1e4 + 5;
  int n, k, lim, c[N], buc[N], pre[N];
  bool ok(int l, int r) {
    for (int i = l; i <= r; i++)
      if (buc[i] == lim) return 0;
    return 1;
  }
  void add(int l, int r) {
    for (int i = l; i <= r; i++) buc[i]++;
  }
  vector<pii> seq;
  pii ans[N];
  int main() {
    cin >> n >> k, lim = (n - 1) / (k - 1) + 1;
    for (int i = 1; i <= n * k; i++) {
      cin >> c[i];
      if (pre[c[i]]) seq.emplace_back((pii){pre[c[i]], i});
      pre[c[i]] = i;
    }
    sort(seq.begin(), seq.end()), reverse(seq.begin(), seq.end());
    for (pii it : seq)
      if (!ans[c[it.first]].first && ok(it.first, it.second))
        add(it.first, it.second), ans[c[it.first]] = it;
    for (int i = 1; i <= n; i++)
      cout << ans[i].first << " " << ans[i].second << endl;
    return 0;
  }
  