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>
const long long mod = 1e8 + 7;
const long long eps = 1e-12;
using namespace std;
int n, a[300000 + 5];
map<int, bool> vis;
vector<pair<int, int> > ans;
int main() {
  while (cin >> n) {
    vis.clear();
    ans.clear();
    for (int(i) = (1); (i) <= (n); ++(i)) cin >> a[i];
    int pre = 0;
    for (int(i) = (1); (i) <= (n); ++(i)) {
      if (vis[a[i]]) {
        ans.push_back(make_pair(pre + 1, i));
        vis.clear();
        pre = i;
      } else
        vis[a[i]] = true;
    }
    if (!ans.size())
      cout << -1 << endl;
    else {
      cout << ans.size() << endl;
      for (int(i) = (0); (i) <= (ans.size() - 1); ++(i))
        cout << ans[i].first << " " << (i == ans.size() - 1 ? n : ans[i].second)
             << endl;
    }
  }
  return 0;
}
