方程的解数

var
  n,m,i,j,t1,t2:longint;
  num,num1,ans:int64;
  tot,tot1:array[0..3385000]of longint;
  k1,k,p,p1:array[0..10000]of longint;
  function max(x,y:int64):int64;
   begin
    if x>y then exit(x) else exit(y);
   end;
    procedure qs(l,r: int64);
      var
         i,j,x,y: int64;
      begin
        i:=l;  j:=r;
         x:=tot[(l+r) div 2];
         repeat
           while tot[i]<x do   inc(i);
           while x<tot[j] do   dec(j);
           if not(i>j) then
             begin
                y:=tot[i]; tot[i]:=tot[j];  tot[j]:=y;
                inc(i);dec(j);
             end;
         until i>j;
         if l<j then    qs(l,j);
         if i<r then    qs(i,r);
      end;
       procedure qs1(l,r: int64);
      var
         i,j,x,y: int64;
      begin
        i:=l;  j:=r;
         x:=tot1[(l+r) div 2];
         repeat
           while tot1[i]<x do   inc(i);
           while x<tot1[j] do   dec(j);
           if not(i>j) then
             begin
                y:=tot1[i]; tot1[i]:=tot1[j];  tot1[j]:=y;
                inc(i);dec(j);
             end;
         until i>j;
         if l<j then    qs1(l,j);
         if i<r then    qs1(i,r);
      end;

 function ksm(a,b:int64):int64;
 var t,y:int64;
  begin
   t:=1; y:=a;
   while b>0 do
    begin
     if (b and 1)=1 then t:=t*y;
     y:=y*y;
     b:=b shr 1;
    end;
    exit(t);
  end;
procedure dfs(now,sum:int64);//求左边未知数的所有解
 var i:longint;
 begin
   if now>n div 2 then begin inc(num);tot[num]:=sum; exit; end;//达到未知数个数了，则记录解
   for i:=1 to m do
    dfs(now+1,sum+k[now]*ksm(i,p[now]));//递归精妙之处，假设有两个未知数 则sum的值是1,1  1,2  2,1   2,2（前属数组省略）能把所有可能记录
 end;
 procedure rdfs(now,sum:int64);//数组和变量最好不要定义太相似，因为错了找错很麻烦，这里我调了好久，比如tot,tot1;  num,num1; 下次注意；
 var i:longint;
 begin
   if now>n-(n div 2) then begin inc(num1);tot1[num1]:=sum; exit; end;
   for i:=1 to m do
    rdfs(now+1,sum+k1[now]*ksm(i,p1[now]));
 end;
begin
 readln(n);
 readln(m);
 for i:=1 to n do
  readln(k[i],p[i]);
  for i:=n div 2+1 to n do
   begin
     inc(j);
     k1[j]:=-1*k[i];
     p1[j]:=p[i];
   end;
  dfs(1,0);//要用搜索，因为不知道一边有几个未知数，for循环不行
  rdfs(1,0);
  qs(1,num);
  qs1(1,num1);
  i:=1;j:=1;
  while (i<=num)and(j<=num1) do
   begin
    if tot[i]=tot1[j] then
    begin//小处理
      t1:=1;t2:=1;
      while (tot[i+1]=tot1[j])and(i<=num) do begin inc(i);inc(t1); end;//i+1是为了让下一位还是原来的数值
      while (tot[i]=tot1[j+1])and(j<=num1) do begin inc(j); inc(t2);end;
      ans:=ans+t1*t2;
      inc(i);inc(j);
    end
    else
     if tot[i]>tot1[j] then inc(j) else inc(i);
   end;
   writeln(ans);
end.