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;
int INF = 1e9;
int const N = 1e6 + 16;
int const M = 1e9 + 7;
int const M2 = 999999983;
int n, k;
int f[N];
int a[N];
pair<int, int> st[N];
inline int add(int x) { return (f[a[x]]++ == 0); }
inline int rem(int x) { return (--f[a[x]] == 0); }
bool chk(int len) {
  int l = 1, r = 1, diff = 0;
  while (r <= len) diff += add(r++);
  bool found = (diff <= k);
  while (!found && r <= n) {
    diff += add(r++);
    diff -= rem(l++);
    found = (diff <= k);
  }
  st[len].first = l;
  st[len].second = r - 1;
  while (l < r) diff -= rem(l++);
  return found;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  int l = min(n, k);
  int r = n;
  int ans = l;
  while (l <= r) {
    int m = l + r >> 1;
    if (chk(m))
      l = m + 1, ans = m;
    else
      r = m - 1;
  }
  printf("%d %d\n", st[ans].first, st[ans].second);
}
