fishery samsung

/*
inputs
5
10
4 5
6 2
10 2
10
8 5
9 1
10 2
24
15 3
20 4
23 7
39
17 8
30 5
31 9
60
57 12
31 19
38 16
outputs
18
25
57
86
339
*/


#include<bits/stdc++.h>
using namespace std;

int pos[4];    // pos represent the position of the slots
int val[4];  // values corresponds to the number of fishermens
int a[1000];

int n, ans = INT_MAX;
/* finding the nearest right empty position*/
int posRight(int index){
    for(int i = index; i<=n; i++)
    if(a[i] == 0)return i;
    return INT_MAX;
}

/*finding the nearest left most empty position*/
int posLeft(int index){
    for(int i = index -1; i>0; i--)
    if(a[i] == 0)return i;
    return INT_MAX;
}


void solve(int x, int y, int z, int currPos, int cost){

  if(cost > ans)return;

  if(val[currPos] == 0){
      if(currPos == x)currPos = y;
      else if(currPos == y)currPos = z;
      else{
          ans = min(ans,cost);
          return;
      }
  }

   int l  = posLeft(pos[currPos]);
   int ll = abs(pos[currPos] - l + 1);
   int r  = posRight(pos[currPos]);
   int rr = abs(r - pos[currPos] + 1);


   int distance = ll>rr ? r : l;
   int tempx = min(ll,rr);

   /* in case of last position if there is equal distance of left and right then check for
      both the cases any of the following will give the optimal results*/

       if(val[currPos] == 1 && ll == rr){

           a[l] = 1;
           val[currPos]--;
           solve(x, y, z, currPos, cost + ll);
           a[l] = 0;
           val[currPos]++;

           a[r] = 1;
           val[currPos]--;
           solve(x, y, z, currPos, cost + rr);
           a[r] = 0;
           val[currPos]++;
       }

   /* optimization is done here we are choosing the pos which is near*/
    else{
       a[distance] = 1;
       val[currPos]--;
       solve(x,y,z,currPos, cost + tempx);
       a[distance] = 0;
       val[currPos]++;
    }
}

void init(){
     for(int i = 0; i<1000; i++)a[i] = 0;
}

int main(){

    int t; cin>>t;
    for(int i = 0; i<t; t--){

        cin>>n;

        for(int i = 1; i<=3; i++)
        cin>>pos[i]>>val[i];

        ans = INT_MAX;
        init();

        solve(1,2,3,1,0);init();
        solve(1,3,2,1,0);init();
        solve(2,3,1,2,0);init();
        solve(2,1,3,2,0);init();
        solve(3,1,2,3,0);init();
        solve(3,2,1,3,0);init();

      cout<<ans<<endl;

    }

    return 0;
}