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;
using ll = long long;
template <class T, class U>
using Pa = pair<T, U>;
template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vec<T>>;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  vec<int> A(N);
  for (int i = 0; i < N; i++) cin >> A[i];
  map<int, int> mp;
  auto del = [&](int x) {
    mp[x]--;
    if (mp[x] == 0) mp.erase(x);
  };
  int r = 0;
  int len = 0;
  int ansl = -1, ansr = -1;
  for (int l = 0; l < N; l++) {
    while (r < N && (mp.size() < K || (mp.size() == K && mp.count(A[r])))) {
      mp[A[r]]++;
      r++;
      if (r - l > len) {
        len = r - l;
        ansl = l, ansr = r;
      }
    }
    del(A[l]);
  }
  cout << ansl + 1 << " " << ansr << "\n";
}
