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
#include <bits/stdc++.h>
using namespace std;
char _;
const int maxN = 300 * 1000 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int base = 701;
const int SQ = 600;
const int maxL = 20;
int a[maxN];
int ted[maxN];
int32_t main() {
  int n;
  scanf("%d", &n);
  vector<int> v;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), v.push_back(a[i]);
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
  vector<int> ans;
  ans.push_back(-1);
  for (int i = 0; i < n; i++) {
    ted[a[i]]++;
    if (ted[a[i]] == 2) {
      int st = ans.back() + 1;
      for (int j = st; j < i + 1; j++) ted[a[j]]--;
      ans.push_back(i);
    }
  }
  if ((int)ans.size() == 1) {
    cout << -1 << endl;
    return 0;
  }
  cout << (int)ans.size() - 1 << endl;
  ans.back() = n - 1;
  for (int i = 1; i < (int)ans.size(); i++)
    cout << ans[i - 1] + 2 << ' ' << ans[i] + 1 << endl;
  return 0;
}
