1 #include<bits/stdc++.h> 2 #define _for(i,a,b) for(register int i = (a);i < b;i ++) 3 #define _rep(i,a,b) for(register int i = (a);i > b;i --) 4 #define INF 0x3f3f3f3f 5 #define MOD 100000000 6 #define maxn 1000003 7 #define pb push_back 8 #define debug() printf("Miku Check OK! ") 9 typedef long long ll; 10 11 using namespace std; 12 13 inline ll read() 14 { 15 ll ans = 0; 16 char ch = getchar(), last = ' '; 17 while(!isdigit(ch)) last = ch, ch = getchar(); 18 while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar(); 19 if(last == '-') ans = -ans; 20 return ans; 21 } 22 inline void write(ll x) 23 { 24 if(x < 0) x = -x, putchar('-'); 25 if(x >= 10) write(x / 10); 26 putchar(x % 10 + '0'); 27 } 28 struct Point 29 { 30 int x; 31 int y; 32 bool operator < (Point b) 33 { 34 return x < b.x; 35 } 36 }; 37 struct Num 38 { 39 int x; 40 int y; 41 }; 42 43 int n,d; 44 Point in[maxn]; 45 Num qMax[maxn]; 46 Num qMin[maxn]; 47 int ans = INT_MAX; 48 bool C(int m) 49 { 50 int frontMax,backMax; 51 int frontMin,backMin; 52 frontMax = 1,backMax = 0; 53 frontMin = 1,backMin = 0; 54 _for(i,1,n+1) 55 { 56 while(frontMax <= backMax && qMax[frontMax].x + m < in[i].x) 57 frontMax ++; 58 59 while(frontMin <= backMin && qMin[frontMin].x + m < in[i].x) 60 frontMin ++; 61 62 while(frontMax <= backMax && qMax[backMax].y <= in[i].y) 63 backMax --; 64 backMax ++; 65 qMax[backMax].y = in[i].y; 66 qMax[backMax].x = in[i].x; 67 68 while(frontMin <= backMin && qMin[backMin].y >= in[i].y) 69 backMin --; 70 backMin ++; 71 qMin[backMin].y = in[i].y; 72 qMin[backMin].x = in[i].x; 73 74 int dd = qMax[frontMax].y-qMin[frontMin].y; 75 if(dd >= d) 76 return true; 77 } 78 return false; 79 } 80 int solve() 81 { 82 int lb = in[1].x,ub = in[n].x+5; 83 while(lb < ub) 84 { 85 int mid = (lb+ub)/2; 86 if(C(mid)) ub = mid; 87 else lb = mid+1; 88 } 89 return lb; 90 } 91 int main() 92 { 93 // freopen("testdata (4).in","r+",stdin); 94 n = read(),d = read(); 95 _for(i,1,n+1) 96 { 97 in[i].x = read(); 98 in[i].y = read(); 99 } 100 101 sort(in+1,in+n+1); 102 ans = solve(); 103 if(ans > in[n].x) 104 write(-1); 105 else 106 write(ans); 107 return 0; 108 }