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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
int inf = 0x3f3f3f3f;
long long infl = 0x3f3f3f3f3f3f3f3fLL;
long double infd = 1.0 / 0.0;
const long long MOD = 1e9 + 7;
const double pi = 2 * acos(0.0);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt = 1;
  while (tt--) {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    map<int, int> mp;
    mp[a[0]]++;
    int x = 1, l = 0, r = 0, as = 1, l1 = 0, rr = 0;
    for (int i = 1; i < n; i++) {
      r++;
      mp[a[i]]++;
      if (mp[a[i]] == 1) x++;
      while (x > k) {
        mp[a[l]]--;
        if (mp[a[l]] == 0) x--;
        l++;
      }
      if (r - l + 1 > as) as = r - l + 1, l1 = l, rr = r;
    }
    cout << l1 + 1 << " " << rr + 1 << '\n';
  }
  return 0;
}
