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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
const long long MOD = 1e9 + 7;
const long long MOD2 = 998244353;
const long long dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long modpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long bpow(long long a, long long b) {
  if (b < 0) {
    return 0;
  }
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long modinv(long long a, long long m = MOD) { return modpow(a, m - 2, m); }
const int N = 5e5 + 5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long r = 0;
    set<long long> s;
    vector<long long> cnt(1e6 + 5, 0);
    long long l, R;
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      r = max(r, i);
      while (r < n && s.size() <= k) {
        if (s.size() == k && s.find(a[r]) == s.end()) {
          break;
        }
        s.insert(a[r]);
        cnt[a[r]]++;
        r++;
      }
      if (s.size() <= k) {
        if (ans < r - i) {
          l = i + 1;
          R = r;
          ans = r - i;
        }
      }
      cnt[a[i]]--;
      if (cnt[a[i]] == 0) {
        s.erase(a[i]);
      }
    }
    cout << l << " " << R;
  }
}
