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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " |       ";
  __f(comma + 1, args...);
}
void FastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(15);
}
long long modpow(long long a, long long p, long long mod) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = (ret * a) % mod;
    a = (a * a) % mod;
    p /= 2;
  }
  return ret % mod;
}
long long power(long long a, long long p) {
  long long ret = 1;
  while (p) {
    if (p & 1) ret = (ret * a);
    a = (a * a);
    p /= 2;
  }
  return ret;
}
vector<pair<int, int> > vec;
int a[300010];
int main() {
  FastIO();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int i = 0, ans = 0;
  while (i < n) {
    map<int, int> cnt;
    int j = i;
    while (j < n) {
      if (cnt[a[j]] == 1) break;
      cnt[a[j]]++;
      j++;
    }
    if (j != n) {
      ans++;
      vec.push_back(make_pair(i, j));
    }
    i = j + 1;
  }
  if (ans == 0) {
    cout << -1 << '\n';
    return 0;
  }
  cout << ans << '\n';
  for (int i = 0; i < (int)(vec.size()) - 1; i++)
    cout << vec[i].first + 1 << " " << vec[i].second + 1 << '\n';
  cout << vec[(int)(vec.size()) - 1].first + 1 << " " << n << '\n';
  return 0;
}
