【题目描述】

Amugae is in a very large round corridor. The corridor consists of two areas. The inner area is equally divided by n sectors, and the outer area is equally divided by m sectors. A wall exists between each pair of sectors of same area (inner or outer), but there is no wall between the inner area and the outer area. A wall always exists at the 12 o'clock position. The inner area's sectors are denoted as (1,1),(1,2),…,(1,n)in clockwise direction. The outer area's sectors are denoted as (2,1),(2,2),…,(2,m)in the same manner. For a clear understanding, see the example image above.Amugae wants to know if he can move from one sector to another sector. He has qquestions.For each question, check if he can move between two given sectors.

Input

The first line contains three integers n, m and q (1≤n,m≤1018, 1≤q≤104) — the number of sectors in the inner area, the number of sectors in the outer area and the number of questions.Each of the next qlines contains four integers sx, sy, ex, ey (1≤sx,ex≤2; if sx=1, then 1≤sy≤n, otherwise 1≤sy≤m; constraints on ey are similar). Amague wants to know if it is possible to move from sector (sx,sy) to sector (ex,ey)

Output

For each question, print "YES" if Amugae can move from (sx,sy)to (ex,ey), and "NO" otherwise.You can print each letter in any case (upper or lower).

Example
Input

4 6 3
1 1 2 3
2 6 1 2
2 6 2 4

Output

YES
NO
YES

【题目分析】
很简单的一道题，就是分类讨论，不过1A还是很爽
【AC代码】

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<climits>
#include<queue>
#include<vector>
#include<set>
#include<map>
using namespace std;typedef long long ll;
const int INF=0x3f3f3f3f;
const int MAXN=1e4+5;
ll n,m,x,y;
int T;ll gcd(ll a,ll b)
{return a%b==0?b:gcd(b,a%b);
}int main()
{ll tmp;ll s1,s2,e1,e2,is,ie,l,r;bool flag;scanf("%lld%lld%d",&n,&m,&T);while(T--){flag=false;scanf("%lld%lld%lld%lld",&s1,&s2,&e1,&e2);tmp=gcd(n,m);x=n/tmp; y=m/tmp;if(s1==e1){if(s1==1){is=(s2%x==0)?(s2/x-1):s2/x;ie=(e2%x==0)?(e2/x-1):e2/x;if(is==ie){flag=true;}}else if(s1==2){is=(s2%y==0)?(s2/y-1):s2/y;ie=(e2%y==0)?(e2/y-1):e2/y;if(is==ie){flag=true;}}}else{if(s1==2){s1=1; e1=2; swap(s2,e2);}if(s2%x==0){r=s2/x*y;l=r-y;}else{l=s2/x*y;r=l+y;}if(e2>l && e2<=r){flag=true;}}if(flag) printf("YES\n");else printf("NO\n");}return 0;
}