There are N pinholes on the xy-plane. The i-th pinhole is located at (x_i,y_i).

We will denote the Manhattan distance between the i-th and j-th pinholes as d(i,j)(=|x_i-x_j|+|y_i-y_j|).

You have a peculiar pair of compasses, called Manhattan Compass. This instrument always points at two of the pinholes. The two legs of the compass are indistinguishable, thus we do not distinguish the following two states: the state where the compass points at the p-th and q-th pinholes, and the state where it points at the q-th and p-th pinholes.

When the compass points at the p-th and q-th pinholes and d(p,q)=d(p,r), one of the legs can be moved so that the compass will point at the p-th and r-th pinholes.

Initially, the compass points at the a-th and b-th pinholes. Find the number of the pairs of pinholes that can be pointed by the compass.

Constraints

* 2≦N≦10^5
* 1≦x_i, y_i≦10^9
* 1≦a < b≦N
* When i ≠ j, (x_i, y_i) ≠ (x_j, y_j)
* x_i and y_i are integers.

Input

The input is given from Standard Input in the following format:


N a b
x_1 y_1
:
x_N y_N


Output

Print the number of the pairs of pinholes that can be pointed by the compass.

Examples

Input

5 1 2
1 1
4 3
6 1
5 5
4 8


Output

4


Input

6 2 3
1 3
5 3
3 5
8 4
4 7
2 5


Output

4


Input

8 1 2
1 5
4 3
8 2
4 7
8 8
3 3
6 6
4 8


Output

7
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}
#define int long long
const int MN=100000;
int inf=1e9;
struct unionfind{
	int par[MN];
	void init(int N){rep(i,N) par[i]=i;}
	int find(int x){
		if(par[x]==x) return x;
		return par[x]=find(par[x]);
	}
	bool same(int x,int y){
		return find(x)==find(y);
	}
	void unite(int x,int y){
		x=find(x),y=find(y);
		if(x==y) return;
		par[y]=x;
	}
}UF;

typedef pair<int,int> P;	//val,id
typedef vector<P> VP;
map<int,VP> mp;
int N,a,b,x[MN],y[MN];
int cnt[MN];

signed main(){
	cin>>N>>a>>b;
	UF.init(N);
	a--,b--;
	rep(i,N) cin>>x[i]>>y[i];
	int D=abs(x[a]-x[b])+abs(y[a]-y[b]);
	rep(i,N){
		mp[x[i]+y[i]].pb(P(x[i]-y[i],i));
	}
	for(auto& it:mp){
		VP& vp=it.sc;
		sort(all(vp));
	}
	for(auto& it:mp){
		VP& vp=it.sc;
		int oh=it.fs;
		rep(i,2){
			int h=(i==0?oh-D:oh+D);
			if(mp.find(h)==mp.end()) continue;
			VP& vs=mp[h];
			int K=vs.size();
			vector<int> done(K+1,0);
			for(P p:vp){
				int x=p.fs;
				int id=p.sc;
				int lid=lower_bound(all(vs),P(x-D,-1))-vs.begin();
				int rid=lower_bound(all(vs),P(x+D,inf))-vs.begin()-1;
				if(lid<=rid){
					UF.unite(id,vs[lid].sc);
					cnt[id]+=rid-lid+1;
					done[lid]++;
					done[rid]--;
				}
			}
			rep(i,K) done[i+1]+=done[i];
			rep(i,K-1) if(done[i]>0) UF.unite(vs[i].sc,vs[i+1].sc);
		}
	}

	mp.clear();

	rep(i,N){
		mp[x[i]-y[i]].pb(P(x[i]+y[i],i));
	}
	for(auto& it:mp){
		VP& vp=it.sc;
		sort(all(vp));
	}
	for(auto& it:mp){
		VP& vp=it.sc;
		int oh=it.fs;
		rep(i,2){
			int h=(i==0?oh-D:oh+D);
			if(mp.find(h)==mp.end()) continue;
			VP& vs=mp[h];
			int K=vs.size();
			vector<int> done(K+1,0);
			for(P p:vp){
				int x=p.fs;
				int id=p.sc;
				int lid=lower_bound(all(vs),P(x-D,-1))-vs.begin();
				int rid=lower_bound(all(vs),P(x+D,inf))-vs.begin()-1;
				if(lid<=rid){
					UF.unite(id,vs[lid].sc);
					done[lid]++;
					done[rid]--;
				}
				lid=lower_bound(all(vs),P(x-D+1,-1))-vs.begin();
				rid=lower_bound(all(vs),P(x+D-1,inf))-vs.begin()-1;
				if(lid<=rid){
					cnt[id]+=rid-lid+1;
				}
			}
			rep(i,K) done[i+1]+=done[i];
			rep(i,K-1) if(done[i]>0) UF.unite(vs[i].sc,vs[i+1].sc);
		}
	}
	int ans=0;
	rep(i,N){
		if(UF.same(i,a)){
			ans+=cnt[i];
		}
	}
	cout<<ans/2<<endl;
}
