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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mcm(long long a, long long b) { return a * b / gcd(a, b); }
long long sq(long long a) { return a * a; }
long long pot(int a, int b) {
  return b ? sq(pot(a, b >> 1)) * (b & 1 ? a : 1) : 1;
}
long long potMod(long long a, long long e) {
  return e ? ((sq(potMod(a, e >> 1)) % 1000000007) * (e & 1 ? a : 1)) %
                 1000000007
           : 1;
}
int cnt[1000006], lo, hi, o, k, n, dif, a[600000], u, tmp;
int main() {
  tmp = -0x3f3f3f3f;
  dif = 0;
  memset(cnt, 0, sizeof(cnt));
  o = 0;
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    if (cnt[a[i]] == 0) dif++;
    cnt[a[i]]++;
    while (dif > k) {
      cnt[a[o]]--;
      if (cnt[a[o]] == 0) dif--;
      o++;
    }
    if (tmp < i - o) {
      tmp = i - o;
      lo = o;
      hi = i;
    }
  }
  cout << lo + 1 << " " << hi + 1 << endl;
}
