Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int inf_int = 2e9;
long long inf_ll = 2e18;
const double pi = 3.1415926535898;
template <typename T>
void prin(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i];
    if (i < a.size() - 1)
      cout << " ";
    else
      cout << "\n";
  }
}
template <typename T>
void prin(set<T>& a) {
  for (auto it = a.begin(); it != a.end(); it++) {
    cout << *it << " ";
  }
}
template <typename T>
void prin_new_line(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << "\n";
  }
}
template <typename T, typename T1>
void prin_new_line(vector<pair<T, T1> >& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
}
int sum_vec(vector<int>& a) {
  int s = 0;
  for (int i = 0; i < a.size(); i++) {
    s += a[i];
  }
  return s;
}
template <typename T>
T max(vector<T>& a) {
  T ans = a[0];
  for (int i = 1; i < a.size(); i++) {
    ans = max(ans, a[i]);
  }
  return ans;
}
template <typename T>
T min(vector<T>& a) {
  T ans = a[0];
  for (int i = 1; i < a.size(); i++) {
    ans = min(ans, a[i]);
  }
  return ans;
}
template <typename T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <typename T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  map<int, int> mp;
  vector<pair<int, int> > ans;
  int cur = 1;
  for (int i = 1; i <= n; i++) {
    int x = a[i];
    if (mp[x] >= cur) {
      ans.push_back({cur, i});
      cur = i + 1;
    }
    mp[x] = i;
  }
  if (cur == 1) {
    cout << -1;
    return;
  }
  if (cur != n + 1) {
    ans[ans.size() - 1].second = n;
  }
  cout << ans.size() << "\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first << " " << ans[i].second << "\n";
  }
}
int main() {
  if (!0) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  int t = 1;
  while (t--) solve();
  return 0;
}