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;
void SIO(string name = "") {
  if (name != "") {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
const long long inf = 1LL << 62LL;
const long long mod = 1e9 + 7;
const long long maxn = 7501;
signed main() {
  SIO("");
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) cin >> a[i];
  long long l = 0, r = 0, i = 0, j = 0;
  multiset<long long> ms;
  set<long long> s;
  while (i <= j && j != n) {
    ms.insert(a[j]);
    s.insert(a[j]);
    while (s.size() > k) {
      ms.erase(ms.find(a[i]));
      if (ms.count(a[i]) == 0) s.erase(a[i]);
      i++;
    }
    if (j - i >= r - l) l = i, r = j;
    j++;
  }
  cout << l + 1 << " " << r + 1 << endl;
}
