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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n = 0;
  cin >> n;
  vector<int> ans;
  int cur = 0;
  map<int, int> am;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    am[a]++;
    cur++;
    if (am[a] > 1) {
      ans.push_back(cur);
      am.clear();
      cur = 0;
    }
  }
  if (cur) {
    if (ans.size() == 0) {
      cout << -1;
      return 0;
    } else
      ans[ans.size() - 1] += cur;
  }
  int last = 1;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << last << ' ' << last + ans[i] - 1 << endl;
    last += ans[i];
  }
  return 0;
}
