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 = 555555;
const int MOD = 1e9 + 7;
const long long INF = 1e18;
int n, m;
map<int, int> f;
int a[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int j = 0, best = -1, _m = 0, be, en;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) {
    f[a[i]]++;
    if (f[a[i]] == 1) _m++;
    while (_m > m) {
      f[a[j]]--;
      if (f[a[j]] == 0) _m--;
      j++;
    }
    if (best < i - j + 1) {
      best = i - j + 1;
      be = j;
      en = i;
    }
  }
  cout << be + 1 << " " << en + 1;
}