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;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
const ll MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
void setIO() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
}
int main() {
  setIO();
  int n, k;
  cin >> n >> k;
  vi a(n);
  for (auto &aa : a) cin >> aa;
  int cur = 0;
  int cnt = 0;
  unordered_map<int, int> m;
  int ans = 0;
  int l = 1, r = 1;
  for (int i = 0; i < (n); ++i) {
    if (m[a[i]] == 0) cnt++;
    m[a[i]]++;
    while (cnt > k) {
      if (m[a[cur]] == 1) cnt--;
      m[a[cur]]--;
      cur++;
    }
    if (i - cur > ans) {
      ans = i - cur;
      l = cur + 1;
      r = i + 1;
    }
  }
  cout << l << ' ' << r << endl;
  return 0;
}