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>
const long long lim = 1e5 + 5;
using namespace std;
long long dx[] = {0, 1, 0, -1};
long long dy[] = {1, 0, -1, 0};
bool vis[lim];
long long n, m, k, x, y, z, ans, cnt, sum;
vector<pair<long long, long long> > adj[100005];
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    cin >> n;
    long long a[n + 3];
    ;
    for (long long i = 1; i < n + 1; i++) cin >> a[i];
    vector<pair<long long, long long> > v;
    long long st = 1;
    map<long long, long long> mp;
    ;
    for (long long i = 1; i < n + 1; i++) {
      if (mp[a[i]]) {
        v.push_back({st, i});
        st = i + 1;
        mp.clear();
      } else
        mp[a[i]] = i;
    }
    if ((long long)v.size() == 0) {
      cout << -1;
      return 0;
    }
    m = (long long)v.size();
    if (v[m - 1].second != n) {
      pair<long long, long long> k = v[m - 1];
      v.pop_back();
      v.push_back({k.first, n});
    }
    cout << (long long)v.size() << "\n";
    ;
    for (long long i = 0; i < (long long)v.size(); i++)
      cout << v[i].first << " " << v[i].second << "\n";
  }
}
