//Knapsack DP is harder than FFT.
#include<unistd.h>
inline char RC(){static char buf[65536],*p=buf,*q=buf;return p==q&&(q=(p=buf)+read(0,buf,65536))==buf?-1:*p++;}
inline int R(){static char c;int a;while((c=RC())<'0');a=c^'0';while((c=RC())>='0')a*=10,a+=c^'0';return a;}
inline void W(int n){char OB[12],OP=0,buf[12],p;if(n==0)OB[OP++]='0';p=0;while(n)buf[p++]='0'+(n%10),n/=10;for(--p;p>=0;--p)OB[OP++]=buf[p];write(1,OB,OP);}

#include<algorithm>
#include<vector>
using namespace std;
typedef pair<int,int> pii;
#define FOR(i,n) for(int i=0;i<(n);++i)
#define ff first
#define ss second

const int N = 3000;
int n, l, w;
pii p[N];
vector<int> v;

int main(){
    n = R(), l = R(), w = R();
    FOR(i,n) p[i].ff = R(), p[i].ss = R(), v.push_back(p[i].ss);
    sort(p, p + n);
    sort(v.begin(), v.end());
    v.resize(unique(v.begin(), v.end()) - v.begin());

    int ans = 0;
    for(int y: v){
        int lptr = 0, rptr = 1, cnt = 0;
        if(p[0].ss >= y and p[0].ss - y <= l) ++cnt;
        for(; lptr < n; ++lptr){
            if(rptr == n) break;
            while(rptr < n and p[rptr].ff - p[lptr].ff <= w){
                if(p[rptr].ss >= y and p[rptr].ss - y <= l) ++cnt;
                ++rptr;
            }
            ans = max(ans, cnt);
            if(p[lptr].ss >= y and p[lptr].ss - y <= l) --cnt;
        }
    }
    W(ans);
    return 0;
}