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 INF = 0x7f7f7f7f7f7f7f7f;
const long long NINF = -INF;
const long long MAXN = 1e+6 + 8;
int dx8[] = {0, 1, 1, 1, 0, -1, -1, -1}, dy8[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dx4[] = {0, 1, 0, -1}, dy4[] = {1, 0, -1, 0};
long long n, k, a[500005], ct[1000001] = {};
long long first(long long x, pair<long long, long long> &p) {
  set<long long> se;
  memset(ct, 0, sizeof(ct));
  long long cnt = 0;
  for (long long i = (0); i <= (x - 1); i++) {
    if (!ct[a[i]]) cnt++;
    ct[a[i]]++;
  }
  if (cnt <= k) {
    p = pair<long long, long long>(0, x - 1);
    return 1;
  }
  for (long long i = 1; i + x - 1 < n; i++) {
    ct[a[i - 1]]--;
    if (!ct[a[i - 1]]) cnt--;
    if (!ct[a[i + x - 1]]) cnt++;
    ct[a[i + x - 1]]++;
    if (cnt <= k) {
      p = pair<long long, long long>(i, i + x - 1);
      return 1;
    }
  }
  return 0;
}
void MAIN() {
  cin >> n >> k;
  for (long long i = (0); i <= (n - 1); i++) cin >> a[i];
  int lo = 1, hi = n, mid;
  pair<long long, long long> ans, p;
  while (lo <= hi) {
    mid = lo + (hi - lo + 1) / 2;
    if (first(mid, p))
      ans = p, lo = mid + 1;
    else
      hi = mid - 1;
  }
  cout << ans.first + 1 << " " << ans.second + 1 << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    MAIN();
  }
}
