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>
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 1;
const long long N = 1e5;
const long long M = 4002;
const int base = 400;
long long n, m, t, ans, k, a[N], b[N], tong, d[N], c[N], lab[N], l[N], r[N],
    st[N * 4], lazy[N * 4];
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
struct edge {
  long long x, y, w;
};
vector<long long> kq, adj[N];
vector<pair<long long, long long> > e;
string s;
long long pw(long long k, long long n) {
  long long total = 1;
  for (; n; n >>= 1) {
    if (n & 1) total = total * k % mod;
    k = k * k % mod;
  }
  return total;
}
template <typename T>
void read(T& x) {
  bool Neg = false;
  char c;
  for (c = getchar(); c < '0' || c > '9'; c = getchar())
    if (c == '-') Neg = !Neg;
  x = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  if (Neg && x > 0) x = -x;
}
bool cmp(const long long& x, const long long& y) { return adj[x] < adj[y]; }
bool ok;
void mx(long long& x, long long y) { x = max(x, y); }
long long findp(long long u) { return lab[u] < 0 ? u : lab[u] = findp(lab[u]); }
void hop(long long u, long long v) {
  if (lab[u] > lab[v]) swap(u, v);
  lab[u] += lab[v];
  lab[v] = u;
}
void down(long long id) {
  if (!lazy[id]) return;
  st[id * 2] += lazy[id];
  st[id * 2 + 1] += lazy[id];
  lazy[id * 2] += lazy[id];
  lazy[id * 2 + 1] += lazy[id];
  lazy[id] = 0;
}
void update(long long id, long long l, long long r, long long u, long long v) {
  if (u <= l && r <= v) {
    ++st[id];
    ++lazy[id];
    return;
  }
  if (u > r || l > v) return;
  down(id);
  long long mid = (l + r) / 2;
  update(id * 2, l, mid, u, v);
  update(id * 2 + 1, mid + 1, r, u, v);
  st[id] = max(st[id * 2], st[id * 2 + 1]);
}
long long get(long long id, long long l, long long r, long long u,
              long long v) {
  if (u <= l && r <= v) return st[id];
  if (u > r || l > v) return 0;
  down(id);
  long long mid = (l + r) / 2;
  return max(get(id * 2, l, mid, u, v), get(id * 2 + 1, mid + 1, r, u, v));
}
void sol() {
  cin >> n >> k;
  priority_queue<pair<pair<long long, long long>, long long>,
                 vector<pair<pair<long long, long long>, long long> >,
                 greater<pair<pair<long long, long long>, long long> > >
      pq;
  for (int i = 1; i <= n * k; i++) {
    cin >> a[i];
    if (b[a[i]]) {
      c[i] = b[a[i]];
      pq.push({{0, i - b[a[i]]}, i});
    }
    b[a[i]] = i;
  }
  while (!pq.empty()) {
    pair<pair<long long, long long>, long long> x = pq.top();
    pq.pop();
    if (x.first.first != get(1, 1, n * k, c[x.second], x.second)) {
      pq.push({{get(1, 1, n * k, c[x.second], x.second), x.first.second},
               x.second});
      continue;
    }
    if (l[a[x.second]] ||
        get(1, 1, n * k, c[x.second], x.second) >= (n + k - 2) / (k - 1))
      continue;
    update(1, 1, n * k, c[x.second], x.second);
    l[a[x.second]] = c[x.second];
    r[a[x.second]] = x.second;
  }
  for (int i = 1; i <= n; i++) cout << l[i] << " " << r[i] << '\n';
}
int main() {
  if (fopen("MARATHON"
            ".inp",
            "r")) {
    freopen(
        "MARATHON"
        ".inp",
        "r", stdin);
    freopen(
        "MARATHON"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int ntest = 1;
  while (ntest-- > 0) sol();
}
