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;
const int N = 300500;
int n, a[N];
vector<int> v;
int cnt[N];
vector<int> all;
int main() {
  scanf("%d", &n);
  for (int i = 1; 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());
  if (n == (int)v.size()) {
    cout << -1;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  }
  vector<pair<int, int> > ans;
  for (int i = 1; i <= n; ++i) {
    int id = -1;
    for (int j = i; j <= n; ++j) {
      if (cnt[a[j]]) {
        id = j;
        break;
      }
      ++cnt[a[j]];
    }
    if (id == -1) {
      ans[(int)ans.size() - 1].second = n;
      break;
    } else {
      ans.push_back(make_pair(i, id));
      for (int j = i; j <= id; ++j) cnt[a[j]] = 0;
      i = id;
    }
  }
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < (int)ans.size(); ++i)
    printf("%d %d\n", ans[i].first, ans[i].second);
  return 0;
}
