The only difference between easy and hard versions is constraints.

You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [l_i; r_i] (l_i ≤ r_i) and it covers all integer points j such that l_i ≤ j ≤ r_i.

The integer point is called bad if it is covered by strictly more than k segments.

Your task is to remove the minimum number of segments so that there are no bad points at all.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 200) — the number of segments and the maximum number of segments by which each integer point can be covered.

The next n lines contain segments. The i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 200) — the endpoints of the i-th segment.

Output

In the first line print one integer m (0 ≤ m ≤ n) — the minimum number of segments you need to remove so that there are no bad points.

In the second line print m distinct integers p_1, p_2, ..., p_m (1 ≤ p_i ≤ n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.

Examples

Input


7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9


Output


3
1 4 7 


Input


5 1
29 30
30 30
29 29
28 30
30 30


Output


3
1 2 4 


Input


6 1
2 3
3 3
2 3
2 2
2 3
2 3


Output


4
1 3 5 6 
#include <iostream>
#include <algorithm>
using namespace std;

int arr2[200], arr2i;

int main() {

    int n, limit;
    
    cin>>n>>limit;
    int arr[n][2];
    
    bool mat[n][n]={};
    
    for ( int i=0; i<n; ++i )
        cin>>arr[i][0]>>arr[i][1];
        
    for( int i= 0; i<n; ++i )
        for( int j=0; j < n; ++j)
            if( (arr[i][0] >= arr[j][0] && arr[i][0] <= arr[j][1]) || (arr[i][1] >= arr[j][0] && arr[i][1] <= arr[j][1]) )
                mat[i][j] = mat[j][i] = 1;
                
    ///////////////////////////////////////////////////////////////////

while (1)
{    
    int max = -1, biggest;
    
    for ( int  i = 0; i < n; ++i )
    {
        int count = 0;
        
        for ( int j = 0; j < n; ++j )
            if ( mat[i][j] == 1 )
                ++count;
                
        if( count > max )
        {
            max = count;
            biggest = i;
        }
    }
    
    if ( max > limit )
        arr2[arr2i++] = biggest + 1;
    else break;
    
    for ( int  i = 0; biggest != -1 && i < n; ++i )
    {
        mat[i][biggest] = 0;
        mat[biggest][i] = 0;
    }
}
    //////////////////////////////////////////////////////////////////
    
    sort( arr2, arr2 +arr2i );
    
    cout<<arr2i<<'\n';
    for ( int i = 0 ; i < arr2i; ++i)
        cout<<arr2[i]<<' ';

    return 0;
}
