/*大水题,用比较笨的方法模拟的*/
#include <stdio.h>
#include <string.h>
int part[200][4000];
int main()
{
int N, time, x, p, count, cal[4000], a, b, i, t;
scanf( "%d", &time );
while( time-- )
{
memset( part, 0, sizeof(part) );
memset( cal, 0, sizeof(cal) );
scanf( "%d", &N );
scanf( "%d", &p );
for( a = 1; a <= p; a++ )
{
t = 0;
scanf( "%d", &x );
for( i = 1; i <= N; i++ )
{
if( ++t == x )
{
part[a][i] = 1;
t = 0;
}
}
}
for( b = 1; b <= N; b++ )
for( i = 1; i <= p; i++ )
cal[b] = part[i][b] | cal[b];
t = 0;
for( i = 1; i <= N; i++ )
{
if( ++t == 6 )
{
cal[i] = 0;
cal[i+1] = 0;
i++;
t = 0;
}
}
count = 0;
for( i = 1; i <= N; i++ )
if( cal[i] )
count++;
printf( "%d\n", count );
}
return 0;
}