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;
const long long int MAXN = 1e6 + 5, INF = 1e18, MOD = 1e9 + 7;
long long int mx = -1e18, mn = 1e18;
long long int n, k, a[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int r = 0, sz = 0, mx = 0, ansl = 0, ansr = 0;
  map<long long int, long long int> mp;
  for (int l = 0; l < n; l++) {
    if (l > r) {
      r = l;
    }
    while (r < n && ((mp[a[r]] == 0 && sz + 1 <= k) || (mp[a[r]] != 0))) {
      if (mp[a[r]] == 0) {
        sz++;
      }
      mp[a[r]]++;
      mx = max(mx, r - l);
      if (mx == r - l) {
        ansl = l;
        ansr = r;
      }
      r++;
    }
    mp[a[l]]--;
    if (mp[a[l]] == 0) {
      sz--;
    }
  }
  cout << ansl + 1 << " " << ansr + 1;
  return 0;
}
