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 MAX_N = 5e5 + 5;
const int M = 1e6;
const long long INF = (long long)(1e18);
const int inf = 1e9;
const long long MOD = 1000000007LL;
int n, k;
int a[MAX_N];
map<int, int> mp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int len = 0;
  int ansL, ansR;
  int l = 0;
  int r = 0;
  while (r < n) {
    while (r < n) {
      if (mp.size() > k) break;
      mp[a[r]]++;
      r++;
    }
    if (r == n && mp.size() <= k) r++;
    if (r - l - 1 > len) {
      len = r - l - 1;
      ansL = l + 1;
      ansR = r - 1;
    }
    mp[a[l]]--;
    if (mp[a[l]] <= 0) mp.erase(mp.find(a[l]));
    l++;
  }
  cout << ansL << " " << ansR;
  return 0;
}