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("O2")
using namespace std;
inline long long input() {
  long long n;
  cin >> n;
  return n;
}
long long pw(long long a, long long b) {
  return (!b ? 1 : (b & 1 ? a * pw(a * a, b / 2) : pw(a * a, b / 2)));
}
const int MAXN = 1e6 + 10;
const int MOD = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 8e18;
int v[MAXN], v2[MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    v[i] = a;
  }
  map<int, int> mp;
  long long ans = 0, r = 0, l = 0;
  pair<int, int> pans;
  pans.first = 0, pans.second = 1;
  for (int i = 0; i < n; i++) {
    v2[r] = v[i];
    r++;
    mp[v[i]]++;
    while (mp.size() > k) {
      mp[v2[l]]--;
      if (mp[v2[l]] == 0) mp.erase(mp.find(v2[l]));
      l++;
    }
    if ((r - l) > ans) {
      ans = (r - l);
      pans.second = i + 1;
      pans.first = pans.second - (r - l) + 1;
    }
  }
  cout << pans.first << " " << pans.second;
  return 0;
}
