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;
#pragma GCC diagnostic ignored "-Wmissing-declarations"
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int& x : a) cin >> x;
  vector<int> d(n + 1, -1);
  vector<int> p(n + 1, -1);
  map<int, int> pr;
  for (int i = 0; i < n; ++i) {
    int val = a[i];
    auto it = pr.find(val);
    d[i + 1] = d[i];
    p[i + 1] = p[i];
    if (it != pr.end()) {
      int nd = 1 + d[it->second];
      if (d[i + 1] < nd) {
        d[i + 1] = nd;
        p[i + 1] = it->second;
      }
      it->second = i;
    } else {
      pr[val] = i;
    }
  }
  if (d[n] < 0) {
    cout << (-1) << '\n';
    exit(0);
  };
  vector<pair<int, int> > res;
  int cur = n;
  while (cur > 0) {
    if (p[cur] < 0) {
      res.back().first = 1;
      break;
    }
    res.push_back({p[cur] + 1, cur});
    cur = p[cur];
  }
  cout << res.size() << '\n';
  for (const auto& r : res) cout << r.first << ' ' << r.second << '\n';
}
