#include<bits/stdc++.h> #define debug(...) fprintf(stderr,VA_ARGS) #define DEBUG printf("Passing [%s] in LINE %d\n",FUNCTION,LINE) #define Debug debug("Passing [%s] in LINE %d\n",FUNCTION,LINE) #define all(x) x.begin(),x.end() #define x first #define y second using namespace std; const double eps=1e-8; const double pi=acos(-1.0); typedef long long ll; typedef pair<int,int> pii; template<class T>int chkmin(T &a,T b){return a>b?a=b,1:0;} template<class T>int chkmax(T &a,T b){return a<b?a=b,1:0;} template<class T>T sqr(T a){return aa;} template<class T>T mmin(T a,T b){return a<b?a:b;} template<class T>T mmax(T a,T b){return a>b?a:b;} template<class T>T aabs(T a){return a<0?-a:a;} template<int a>int cmp_a(int x,int y){return a[x]<a[y];} #define min mmin #define max mmax #define abs aabs int read(){ int s=0,base=1; char c; while(!isdigit(c=getchar()))if(c=='-')base=-base; while(isdigit(c)){s=s10+(c^48);c=getchar();} return sbase; } bool cx[1048575]; int mn[1048575],mx[1048575]; template<const function<int(int,int)> &cmp> struct MQ{ int s[1048575],*l,*r; MQ(){l=s+1;r=s;} void add(int x){ while(l<=r && cmp(x,*r))--r; *(++r)=x; } void del(int x){ if(*lx)++l; } int top(){return *l;} }; int increase(int a,int b){return a<b;} int decrease(int a,int b){return a>b;} function<int(int,int)> Inc=increase; function<int(int,int)> Dec=decrease; MQ<Inc> inc; MQ<Dec> dec_; int main(){ #ifdef cnyali_lk freopen("2698.in","r",stdin); freopen("2698.out","w",stdout); #endif int n,d,x,y,l,r,xmn=0x3f3f3f3f; n=read();d=read(); for(int i=0;i<=1000000;++i)mn[i]=0x3f3f3f3f,mx[i]=-0x3f3f3f3f; for(int i=1;i<=n;++i){ x=read();y=read(); chkmax(mx[x],y); chkmin(mn[x],y); cx[x]=1; chkmin(xmn,x); } l=xmn,r=xmn; inc.add(mn[l]); dec_.add(mx[l]); int ans=0x3f3f3f3f; n=1000000; while(r<=n){ while(r<=n && dec_.top()-inc.top()<d){++r;if(cx[r]){inc.add(mn[r]);dec_.add(mx[r]);}} while(l<r && dec_.top()-inc.top()>=d){if(cx[l]){inc.del(mn[l]);dec_.del(mx[l]);}++l;} if(r<=n) chkmin(ans,r-l+1); } if(ans0x3f3f3f3f)printf("-1\n"); else printf("%d\n",ans); return 0; }

#include<bits/stdc++.h>
using namespace std;
void read(int &x){//快读 
	int f=1;x=0;
	char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9') { x=x*10+ch-'0';ch=getchar();}
	x*=f;
}
#define N 100010
int n,d,ans;
struct node{
	int x,y;
}a[N];
bool cmp(node aa,node bb){
	return aa.x<bb.x;
}
int p1[N],p2[N];//我写单调队列一般用 q[] 存具体值，用 p[]存位置 
int head1,head2,tail1,tail2;//队列 1存最大值，队列 2存最小值 
int main(){
	read(n),read(d);//如题所示 
	for(int i=1;i<=n;i++){
		read(a[i].x),read(a[i].y);//结构体，因为要排序 
	}
	sort(a+1,a+1+n,cmp);//排序 
	ans=0x3f3f3f3f;
	head1=head2=1;
	for(int le=0,r=0;le<=n;le++){//最外层循环拉左端点 
		while(head1<=tail1&&p1[head1]<le) head1++;//如果队首比左端点小，那么右移
		while(head2<=tail2&&p2[head2]<le) head2++;//因为排序过了，编号小x一定小 
		while(a[p1[head1]].y-a[p2[head2]].y< d && r<n){
			r++;
			while(head1<=tail1&&a[p1[tail1]].y<a[r].y) tail1--;
			p1[++tail1]=r;
			while(head2<=tail2&&a[p2[tail2]].y>a[r].y) tail2--;
			p2[++tail2]=r;//更新两个单调队列 
		}
		if(r!=n) ans=min(ans,a[r].x-a[le].x);//满足条件 
	} 
	if(ans>=0x3f3f3f3f){//判无解 
		printf("-1");return 0;
	}
	else{
		printf("%d",ans);
	}
}

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