M = m[1], R = r[1]For i = 2 .. N 	d = gcd(M, m[i])	c = r[i] - R	If (c mod d) Then	// 无解的情况		Return -1	End If	(k1, k2) = extend_gcd(M / d, m[i] / d)	// 扩展欧几里德计算k1,k2	k1 = (c / d * k1) mod (m[i] / d)	// 扩展解系	R = R + k1 * M		// 计算x = m[1] * k[1] + r[1]	M = M / d * m[i] 	// 求解lcm(M, m[i])	R %= M 			// 求解合并后的新R，同时让R最小End For		If (R < 0) Then 	R = R + MEnd IfReturn R

#include<cstdio>#include<utility>#include<algorithm>#include<iostream>using namespace std;typedef long long LL;#define mpr make_pairconst int N = 1005;LL n,a1,a2,b1,b2;pair< LL,LL > m[N];inline LL in(LL x=0,char ch=getchar()){ while(ch>‘9‘ || ch<‘0‘) ch=getchar();	while(ch>=‘0‘ && ch<=‘9‘) x=(x<<3)+(x<<1)+ch-‘0‘,ch=getchar();return x; }LL Exgcd(LL a,LL b,LL &x,LL &y){	if(!b){ x=1,y=0;return a; }	LL r=Exgcd(b,a%b,x,y);LL t=x;	x=y,y=t-(a/b)*y;return r;}int Solve(){	LL x,y,d=Exgcd(a1,a2,x,y);	if((b2-b1)%d) return 0;	Exgcd(a1/d,a2/d,x,y),x*=(b2-b1)/d,x=(x%(a2/d)+a2/d)%(a2/d);	b1=a1*x+b1,a1=a1/d*a2,b1=(b1%a1+a1)%a1;	return 1;}int main(){	n=in();	for(LL i=1,u,v;i<=n;i++) u=in(),v=in(),m[i]=mpr(u,v);	a1=m[1].first,b1=m[1].second;	for(int i=2;i<=n;i++){		a2=m[i].first,b2=m[i].second;		if(!Solve()) return puts("-1"),0;	}return printf("%lld\n",b1),0;}