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 void umax(T &x, T y) {
  if (y > x) x = y;
}
template <class T>
inline void umin(T &x, T y) {
  if (y < x) x = y;
}
const int N = 3e5 + 5, B = 0x7fffffff;
int a[N];
int n, pre = 1;
multiset<int> s;
vector<pair<int, int> > res;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    s.insert(a[i]);
    if (s.count(a[i]) == 2) {
      s.clear();
      res.push_back({pre, i});
      pre = i + 1;
    }
  }
  if (res.empty())
    puts("-1");
  else {
    cout << (int)res.size() << "\n";
    res[(int)res.size() - 1].second = n;
    for (auto x : res) cout << x.first << " " << x.second << "\n";
  }
  return !1;
}
