Problem Description

There is a pile of n wooden sticks. The length and weight of each stick are known in advance. The sticks are to be processed by a woodworking machine in one by one fashion. It needs some time, called setup time, for the machine to prepare processing a stick. The setup times are associated with cleaning operations and changing tools and shapes in the machine. The setup times of the woodworking machine are given as follows:

(a) The setup time for the first wooden stick is 1 minute.
(b) Right after processing a stick of length l and weight w , the machine will need no setup time for a stick of length l' and weight w' if l<=l' and w<=w'. Otherwise, it will need 1 minute for setup.

You are to find the minimum setup time to process a given pile of n wooden sticks. For example, if you have five sticks whose pairs of length and weight are (4,9), (5,2), (2,1), (3,5), and (1,4), then the minimum setup time should be 2 minutes since there is a sequence of pairs (1,4), (3,5), (4,9), (2,1), (5,2).

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case consists of two lines: The first line has an integer n , 1<=n<=5000, that represents the number of wooden sticks in the test case, and the second line contains n 2 positive integers l1, w1, l2, w2, ..., ln, wn, each of magnitude at most 10000 , where li and wi are the length and weight of the i th wooden stick, respectively. The 2n integers are delimited by one or more spaces.

Output

The output should contain the minimum setup time in minutes, one per line.

Sample Input

3
5
4 9 5 2 2 1 3 5 1 4
3
2 2 1 1 2 2
3
1 3 2 2 3 1

Sample Output

2
1
3

#include <stdio.h>
#include <stdlib.h>
#include <string.h>int T;struct E
{int len, wi;
} e[5010];char hash[5010];int cmp( const void *a, const void *b )
{if( ( ( struct E * )a )-> len!= ( ( struct E * )b )-> len ){return ( ( struct E * )a )-> len- ( ( struct E * )b )-> len;}else{return ( ( struct E * )a )-> wi- ( ( struct E * )b )-> wi;}
}int find( int N )
{for( int i= 0; i< N; ++i  ){if( !hash[i] ){return i;}}return -1;
}bool ok( int i, int baselen, int basewi )
{if( e[i].wi>= basewi ){return true;}else{return false;}
}int main(  )
{scanf( "%d", &T );while( T-- ){memset( hash, 0, sizeof( hash ) );int N, time= 0;scanf( "%d", &N );for( int i= 0; i< N; ++i ){scanf( "%d %d", &e[i].len, &e[i].wi );}qsort( e, N, sizeof( e[0] ), cmp );int sta;while( 1 ){sta= find( N );//  printf( "sta=%d\n", sta );if( sta== -1 ){break;}int baselen= e[sta].len, basewi= e[sta].wi;//    printf( "blen= %d, bwi= %d\n", baselen, basewi );for( int i= sta; i< N; ++i ){//printf( "blen= %d, bwi= %d\n", baselen, basewi );if( !hash[i]&& ok(i, baselen, basewi ) ){hash[i]= 1;//printf( "..%d  ..%d\n", e[i].len, e[i].wi );baselen= e[i].len;basewi= e[i].wi;}}++time;}printf( "%d\n", time );}
}/*3
5
4 9 5 2 2 1 3 5 1 4
3
2 2 1 1 2 2
3
1 3 2 2 3 1*/