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 INF = 1e18 + 5;
struct node {
  int l, r;
  long long minv;
  node() { minv = INF; }
} seg[4000005];
void build(int id, int l, int r) {
  seg[id].l = l, seg[id].r = r;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(id << 1, l, mid), build(id << 1 | 1, mid + 1, r);
}
void pushup(int id) {
  seg[id].minv = min(seg[id << 1].minv, seg[id << 1 | 1].minv);
}
void update(int id, int x, long long v) {
  if (seg[id].l == seg[id].r && seg[id].l == x) {
    seg[id].minv = v;
    return;
  }
  int mid = (seg[id].l + seg[id].r) >> 1;
  if (x <= mid)
    update(id << 1, x, v);
  else
    update(id << 1 | 1, x, v);
  pushup(id);
}
long long query(int id, int x, int y) {
  if (x <= seg[id].l && seg[id].r <= y) return seg[id].minv;
  int mid = (seg[id].l + seg[id].r) >> 1;
  long long ans = INF;
  if (x <= mid) ans = query(id << 1, x, y);
  if (mid + 1 <= y) ans = min(ans, query(id << 1 | 1, x, y));
  return ans;
}
long long f[1000005];
struct tmp {
  int l, r, w;
} g[300005];
bool cmp(tmp a, tmp b) { return a.l < b.l; }
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  build(1, 1, m);
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d", &g[i].l, &g[i].r, &g[i].w);
  }
  sort(g + 1, g + n + 1, cmp);
  f[m] = 0;
  update(1, m, 0);
  int t = n + 1;
  for (int i = m - 1; i >= 1; i--) {
    f[i] = INF;
    while (t - 1 && g[t - 1].l > i) t--;
    if (t == 1 || g[t - 1].l != i) continue;
    while (t - 1 && g[t - 1].l == i) {
      t--;
      int j = g[t].r;
      f[i] = min(f[i], query(1, i + 1, j) + g[t].w);
    }
    update(1, i, f[i]);
  }
  printf("%lld\n", f[1]);
  return 0;
}
