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 long long MIN = 1e3 + 2;
const long long MXN = 1e6 + 3;
const long long INF = 1e9 + 7;
const long long LINF = 1e18 + 15;
const long double EPS = 1e-9;
long long n, a[MXN], ans, l = 1, c[MXN], d[MXN], e;
set<long long> s;
map<long long, long long> used;
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  if (n - s.size() == 0) {
    cout << -1;
    return 0;
  }
  for (long long i = 1; i <= n; i++) {
    used[a[i]]++;
    if (used[a[i]] == 2) {
      ans++;
      c[ans] = l;
      d[ans] = i;
      l = i + 1;
      for (long long j = d[ans]; j >= c[ans]; j--) {
        used[a[j]] = 0;
      }
    }
  }
  if (d[ans] != n) {
    d[ans] = n;
  }
  cout << ans << "\n";
  for (long long i = 1; i <= ans; i++) {
    cout << c[i] << " " << d[i] << "\n";
  }
  return 0;
}
