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;
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:336777216")
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long lnxt() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
const int base = (int)1e9;
const int N = (int)1e6 + 100;
const int MAXN = (int)1e3 + 100;
const int MOD = (int)1e9 + 7;
const int INF = INT_MAX;
const long long LL_INF = LLONG_MAX;
const long double PI = acos((long double)-1.0);
const long double EPS = 1e-8;
long long cnt = 0, ans = 0;
void solve() {
  int n;
  cin >> n;
  vector<int> V(n);
  bool any = false;
  set<int> S;
  vector<pair<int, int> > Ans;
  int prev = 0;
  int cur;
  for (int i = 0; i < n; ++i) {
    cin >> cur;
    set<int>::iterator it = S.find(cur);
    S.insert(cur);
    if (it != S.end()) {
      Ans.push_back(make_pair(prev + 1, i + 1));
      prev = i + 1;
      S.clear();
      any = true;
    }
  }
  if (!Ans.empty() && Ans[Ans.size() - 1].second != n)
    Ans[Ans.size() - 1].second = n;
  if (!any || Ans[Ans.size() - 1].second != n) {
    cout << -1 << endl;
    return;
  }
  cout << Ans.size() << endl;
  for (auto to : Ans) cout << to.first << ' ' << to.second << endl;
}
int main() {
  int T = 1;
  for (; T; --T) solve();
  return 0;
}