CF 315A(Sereja and Bottles-开易拉罐)

A. Sereja and Bottles

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Sereja and his friends went to a picnic. The guys had n soda bottles just for it. Sereja forgot the bottle opener as usual, so the guys had to come up with another way to open bottles.

Sereja knows that the i-th bottle is from brand a_i, besides, you can use it to open other bottles of brand b_i. You can use one bottle to open multiple other bottles. Sereja can open bottle with opened bottle or closed bottle.

Knowing this, Sereja wants to find out the number of bottles they've got that they won't be able to open in any way. Help him and find this number.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of bottles. The next n lines contain the bottles' description. The i-th line contains two integers a_i, b_i (1 ≤ a_i, b_i ≤ 1000) — the description of the i-th bottle.

Output

In a single line print a single integer — the answer to the problem.

Sample test(s)

input

4
1 1
2 2
3 3
4 4

output

4

input

4
1 2
2 3
3 4
4 1

output

0

水题，直接数。注意b[i]对a[j]。

考场上有一大堆人这题各种卖萌，专业Hack30年……

#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define MAXN (100+10)
#define MAXM (1000+10)
int a[MAXN],b[MAXN],n;
bool b2[MAXN]={0};
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
    memset(b2,0,sizeof(b2));
    cin>>n;
    For(i,n) cin>>a[i]>>b[i];
    For(i,n)
        For(j,n)
        {
            if (i!=j&&b[i]==a[j]) b2[j]=1;
        }
    int ans=0;
    For(i,n) ans+=b2[i];
    cout<