Problem: The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Examples

Input

5 5
1 2 3 4 5


Output

1 5


Input

9 3
6 5 1 2 3 2 1 4 5


Output

3 7


Input

3 1
1 2 3


Output

1 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 3 * 1e5 + 100, M = 3 * 1e5 + 10;
map<int, int> mp;
bool en[N];
int a[N];
int l[N], r[N];
int main() {
  int las = 0;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int lasen = -1;
  for (int i = 0; i < n; i++) {
    if (!mp.count(a[i])) {
      mp[a[i]] = i;
      continue;
    }
    if (lasen < mp[a[i]]) {
      lasen = i;
      en[i] = true;
      mp.erase(a[i]);
    } else {
      mp[a[i]] = i;
    }
  }
  if (lasen == -1) {
    printf("-1\n");
    return 0;
  }
  int sz = 0;
  int L = 0;
  for (int i = 0; i < n; i++) {
    if (en[i]) {
      r[sz] = i;
      l[sz++] = L;
      L = i + 1;
    }
  }
  r[sz - 1] = n - 1;
  printf("%d\n", sz);
  for (int i = 0; i < sz; i++) printf("%d %d\n", l[i] + 1, r[i] + 1);
}