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;
int N, K;
vector<int> vi;
int cnt[1000000 + 10];
int jud(int m) {
  int l, r;
  int res = -1;
  int i, j, k;
  l = r = k = 0;
  for (i = 0; i < N; i++) {
    if (cnt[vi[i]] == 0)
      k++, cnt[vi[i]]++;
    else
      cnt[vi[i]]++;
    r++;
    if (k > K) {
      for (j = l; j < r; j++) {
        cnt[vi[j]]--;
        if (cnt[vi[j]] == 0) k--;
        l++;
        if (k <= K) break;
      }
    }
    if (r - l >= m) res = l;
  }
  if (r - l >= m && k <= K) res = l;
  for (i = l; i < N; i++) cnt[vi[i]]--;
  return res;
}
int main() {
  int i, j, k;
  int n, m;
  cin >> N >> K;
  vi.resize(N);
  for (i = 0; i < N; i++) {
    scanf("%d", &vi[i]);
  }
  int l, r;
  int L, R;
  l = 0, r = N;
  while (l < r) {
    int tl, tr;
    m = (l + r + 1) / 2;
    tl = jud(m);
    if (tl != -1)
      l = m, L = tl, R = tl + m - 1;
    else
      r = m - 1;
  }
  cout << L + 1 << " " << R + 1 << endl;
  return 0;
}
