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 int N = 5e5 + 5;
int n, k;
int a[N];
int c[1000000 + 10];
vector<pair<int, int> > ans;
int num;
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  int l = 0, r = 0;
  num = 0;
  for (l = 0; l < n; ++l) {
    if (l)
      while (num >= k) {
        c[a[l - 1]]--;
        if (c[a[l - 1]] == 0) num--;
        if (num < k) break;
        l++;
      }
    while ((r < n) && (num < k + 1)) {
      if (c[a[r]] == 0) num++;
      c[a[r]]++;
      r++;
    }
    if (r == n) {
      if (num == k + 1) ans.push_back(make_pair(l, r - 2));
      if (num < k + 1) ans.push_back(make_pair(l, r - 1));
      break;
    }
    r--;
    c[a[r]]--;
    num--;
    ans.push_back(make_pair(l, r - 1));
  }
  pair<int, int> last;
  last = ans[0];
  for (int i = 1; i < ans.size(); ++i)
    if ((last.second - last.first) < (ans[i].second - ans[i].first))
      last = ans[i];
  cout << last.first + 1 << " " << last.second + 1;
}
