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;
template <class T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
inline T LCM(T a, T b) {
  if (a < 0) return LCM(-a, b);
  if (b < 0) return LCM(a, -b);
  return a * (b / gcd(a, b));
}
vector<int> arr;
int main() {
  vector<pair<int, int> > ans;
  int n;
  scanf("%d", &n);
  int i, temp;
  for (i = (int)0; i < (int)n; ++i) {
    scanf("%d", &temp);
    arr.push_back(temp);
  }
  set<int> a;
  int start = 0;
  for (i = (int)0; i < (int)n; ++i) {
    if (a.find(arr[i]) != a.end()) {
      a.clear();
      ans.push_back(make_pair(start + 1, i + 1));
      start = i + 1;
    } else {
      a.insert(arr[i]);
    }
  }
  if (a.size() > 0 && ans.size() == 0) {
    printf("-1");
    return 0;
  }
  if (a.size() > 0) {
    ans[ans.size() - 1].second = arr.size();
  }
  cout << ans.size() << endl;
  for (i = (int)0; i < (int)ans.size(); ++i) {
    printf("%d %d\n", ans[i].first, ans[i].second);
  }
  return 0;
}
