(* Enter your code here. Read input from STDIN. Print output to STDOUT *){$mode objfpc}
uses math;

var
  t,n,m:integer;
  a,b,k,f,res:array[0..100000] of integer;
  it,lazy:array[1..2,0..400000] of integer;

procedure sort(l,r: longint);
      var
         i,j,x,y: longint;
      begin
         i:=l;
         j:=r;
         x:=b[random(r-l+1)+l];
         repeat
           while b[i]<x do inc(i);
           while x<b[j] do dec(j);
           if not(i>j) then
             begin
                y:=a[i]; a[i]:=a[j]; a[j]:=y;
                y:=b[i]; b[i]:=b[j]; b[j]:=y;
                y:=k[i]; k[i]:=k[j]; k[j]:=y;
                inc(i); dec(j);
             end;
         until i>j;
         if l<j then
           sort(l,j);
         if i<r then
           sort(i,r);
      end;

procedure down(k,xo:integer);
  begin
    it[k,xo*2]:=it[k,xo*2] + lazy[k,xo];
    lazy[k,xo*2]:=lazy[k,xo*2] + lazy[k,xo];
    it[k,xo*2+1]:=it[k,xo*2+1] + lazy[k,xo];
    lazy[k,xo*2+1]:=lazy[k,xo*2+1] + lazy[k,xo];
    lazy[k,xo]:=0;
  end;

procedure increase(k,u,v,i,j,xo:integer);
var mid:integer;
  begin
    if (v < i) or (u > j) then exit;
    if (u <= i) and (j <= v) then
      begin
        inc(it[k,xo]);
        inc(lazy[k,xo]);
        exit;
      end;
    mid:=(i+j) div 2;
    down(k,xo);
    increase(k,u,v,i,mid,xo*2);
    increase(k,u,v,mid+1,j,xo*2+1);
    it[k,xo]:=max(it[k,xo*2] , it[k,xo*2+1]);
  end;

function query(u,v,i,j,xo:integer):integer;
var mid:integer;
  begin
    if (v < i) or (u > j) then exit(0);
    if (u <= i) and (j <= v) then exit(max(it[1,xo] , it[2,xo]));
    mid:=(i+j) div 2;
    down(1,xo);
    down(2,xo);
    result:=max(query(u,v,i,mid,xo*2) , query(u,v,mid+1,j,xo*2+1));
    it[1,xo]:=max(it[1,xo*2] , it[1,xo*2+1]);
    it[2,xo]:=max(it[2,xo*2] , it[2,xo*2+1]);
  end;

procedure update(o,w,i,j,xo:integer);
var mid:integer;
  begin
    if i = j then
      begin
        it[1,xo]:=w;
        it[2,xo]:=w;
        exit;
      end;
    mid:=(i+j) div 2;
    down(1,xo);
    down(2,xo);
    if o <= mid then update(o,w,i,mid,xo*2)
    else update(o,w,mid+1,j,xo*2+1);
    it[1,xo]:=max(it[1,xo*2] , it[1,xo*2+1]);
    it[2,xo]:=max(it[2,xo*2] , it[2,xo*2+1]);
  end;

procedure solve;
var i,j:integer; o:char;
  begin
    fillchar(it,sizeof(it),0);
    fillchar(lazy,sizeof(lazy),0);
    readln(m,n);
    for i:=1 to n do res[i]:=m+1;
    for i:=1 to n do
      begin
        read(o);
        if (o = 'E') or (o = 'C') then k[i]:=1
        else k[i]:=2;
        if i < n then read(o);
      end;
    readln;
    for i:=1 to n do read(a[i]);
    for i:=1 to n do read(b[i]);
    sort(1,n);
    //for i:=1 to n do writeln(k[i],' ',a[i],' ',b[i]);
    j:=1;
    for i:=1 to m do
      begin
        while (j <= n) and (b[j] <= i) do
          begin
            if a[j] <= b[j] then increase(k[j],1,a[j],1,m,1);
            inc(j);
          end;
        f[i]:=query(1,i,1,m,1);
        ///if i = 5 then writeln(f[i],' ',it[2,1]);
        res[f[i]]:=min(res[f[i]] , i);
        update(i,f[i],1,m,1);
      end;
    for i:=n-1 downto 1 do res[i]:=min(res[i] , res[i+1]);
    for i:=1 to n do
      if res[i] = m+1 then write('-1 ') else write(res[i],' ');
    writeln;
  end;

begin           {
  assign(input,'a.inp');
  reset(input);}
  read(t);
  while t > 0 do
    begin
      dec(t);
      solve;
    end;
end.