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 pii = pair<int, int>;
using pll = pair<ll, ll>;
constexpr int INF = 1e9;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  int lb = 0, ub = n + 1;
  int l = -1, r = -1;
  while (ub - lb > 1) {
    int len = (lb + ub) / 2;
    vector<int> cnt(1000001);
    int sz = 0;
    for (int i = 0; i < len; ++i) {
      cnt[a[i]]++;
      sz += cnt[a[i]] == 1;
    }
    if (sz <= k) {
      l = 1, r = len;
      lb = len;
      continue;
    }
    for (int i = 0; i + len < n; ++i) {
      cnt[a[i]]--;
      sz -= cnt[a[i]] == 0;
      cnt[a[i + len]]++;
      sz += cnt[a[i + len]] == 1;
      if (sz <= k) {
        l = i + 2, r = i + len + 1;
        break;
      }
    }
    if (r - l + 1 == len) {
      lb = len;
    } else {
      ub = len;
    }
  }
  cout << l << ' ' << r << endl;
}