program xibao;{细胞个数}
const dx:array[1..4] of -1..1=(-1,0,1,0);
      dy:array[1..4] of -1..1=(0,1,0,-1);
var int:text;
    name,s:string;
    pic:array[1..50,1..79] of byte;
    bz:array[1..50,1..79] of boolean;
    m,n,i,j,num:integer;
    h:array[1..4000,1..2] of byte;
procedure doing(p,q:integer);
  var i,t,w,x,y:integer;
  begin
    inc(num);bz[p,q]:=false;
    t:=1;w:=1;h[1,1]:=p;h[1,2]:=q;{遇到的第一个细胞入队}
    repeat
      for i:=1 to 4 do{沿细胞的上下左右四个方向搜索细胞}
        begin
          x:=h[t,1]+dx[i];y:=h[t,2]+dy[i];
          if (x>0) and (x<=m) and (y>0) and (y<=n) and bz[x,y]
          then begin inc(w);h[w,1]:=x;h[w,2]:=y;bz[x,y]:=false;end;{为细胞的入队}
end;
inc(t);{队头指针加1}
until t>w;{直至队空为止}
end;
begin
fillchar(bz,sizeof(bz),true);num:=0;
assign(input,'b.in');
assign(output,'b.out');
reset(input);
rewrite(output);
readln(m,n);
for i:=1 to m do
begin readln(s);
for j:=1 to n do
begin pic[i,j]:=ord(s[j])-ord('0');
if pic[i,j]=0 then bz[i,j]:=false;
end;
end;
for i:=1 to m do
for j:=1 to n do if bz[i,j] then doing(i,j);{在矩阵中寻找细胞}
writeln(num);
close(input);
close(output);
end.