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 = 500005;
int n, k, a[N], used[2 * N], cnt, L, R, l, r, m;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  l = 1;
  r = n;
  while (l <= r) {
    cnt = 0;
    m = (l + r) / 2;
    pair<int, int> mn;
    for (int i = 1; i <= m; i++) {
      used[a[i]]++;
      cnt += (used[a[i]] == 1);
    }
    mn = make_pair(cnt, m);
    for (int i = m + 1; i <= n; i++) {
      used[a[i]]++;
      cnt += (used[a[i]] == 1);
      used[a[i - m]]--;
      cnt -= (used[a[i - m]] == 0);
      mn = min(make_pair(cnt, i), mn);
    }
    for (int i = 1; i <= n; i++) used[a[i]] = 0;
    if (mn.first <= k) {
      R = mn.second;
      L = R + 1 - m;
      l = m + 1;
    } else
      r = m - 1;
  }
  cout << L << " " << R << endl;
  return 0;
}
