Algorithm

D. Longest k-Good Segment
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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
Copy
`5 51 2 3 4 5`
output
Copy
`1 5`
input
Copy
`9 36 5 1 2 3 2 1 4 5`
output
Copy
`3 7`
input
Copy
`3 11 2 3`
output
Copy
`1 1`

Code Examples

#1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <stdio.h>
#include <queue>
#include <algorithm>

using namespace std;

int const N = 5e5 + 1;
int a[N], freq[2 * N];
queue<int> q;

int main() {
int n, m;
scanf("%d%d", &n, &m);

for(int i = 0; i < n; i++)
scanf("%d", &a[i]);

int un = 0, res = 0, r, l;

for(int i = 0; i < n; i++) {
if(freq[a[i]]++ == 0)
un++;

q.push(a[i]);

while(!q.empty() && un > m) {
if(--freq[q.front()] == 0)
un--;

q.pop();
}

if(res < (int)q.size()) {
res = q.size();
l = i - res + 1;
r = i;
}
}

printf("%d %d\n", l + 1, r + 1);

return 0;
}``````
Copy The Code &

Input

cmd
5 5
1 2 3 4 5

Output

cmd
1 5