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", "-funroll-all-loops")
using namespace std;
const int N = 1e6 + 10;
int n, k, a[N], res, l, r, cnt[N], s, ql, qr;
inline void add(int x) {
  if (++cnt[x] == 1) s++;
}
inline void del(int x) {
  if (--cnt[x] == 0) s--;
}
signed main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  l = r = 1, add(a[l]);
  while (l <= r && r <= n) {
    while (s > k) del(a[l++]);
    if (r - l + 1 > res) res = r - l + 1, ql = l, qr = r;
    add(a[++r]);
  }
  cout << ql << ' ' << qr;
  return 0;
}
