Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:134217728")
using namespace std;
const int N = 5e5 + 100;
int a[N];
int d[(int)1e6 + 100];
int n, k;
pair<int, int> solve() {
  int c = 0;
  int r = 0;
  pair<int, int> ans = make_pair(0, 0);
  for (int l = 0; l < n; l++) {
    while (r < n) {
      if (c == k && d[a[r]] == 0) break;
      d[a[r]]++;
      if (d[a[r]] == 1) c++;
      r++;
    }
    if (ans.second - ans.first + 1 < r - l) ans = make_pair(l, r - 1);
    d[a[l]]--;
    if (d[a[l]] == 0) c--;
  }
  return ans;
}
pair<int, int> tupoe() {
  pair<int, int> ans = make_pair(0, 0);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      set<int> S;
      for (int k = i; k <= j; k++) S.insert(a[k]);
      if (S.size() <= k) {
        if (ans.second - ans.first < j - i) ans = make_pair(i, j);
      }
    }
  }
  return ans;
}
void gen() {
  n = 14;
  k = 4;
  for (int i = 0; i < n; i++) {
    a[i] = rand() % 6;
  }
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  pair<int, int> ans = solve();
  cout << ans.first + 1 << ' ' << ans.second + 1;
  return 0;
}