codeforces 720A：Closing ceremony

Description

The closing ceremony of Squanch Code Cup is held in the big hall with n × m seats, arranged in n rows, m seats in a row. Each seat has two coordinates (x, y) (1 ≤ x ≤ n, 1 ≤ y ≤ m).

There are two queues of people waiting to enter the hall: k people are standing at (0, 0) and n·m - k people are standing at (0, m + 1). Each person should have a ticket for a specific seat. If person p at (x, y) has ticket for seat (x_p, y_p) then he should walk |x - x_p| + |y - y_p| to get to his seat.

Each person has a stamina — the maximum distance, that the person agrees to walk. You should find out if this is possible to distribute all n·m tickets in such a way that each person has enough stamina to get to their seat.

Input

The first line of input contains two integers n and m (1 ≤ n·m ≤ 10⁴) — the size of the hall.

The second line contains several integers. The first integer k (0 ≤ k ≤ n·m) — the number of people at (0, 0). The following k integers indicate stamina of each person there.

The third line also contains several integers. The first integer l (l = n·m - k) — the number of people at (0, m + 1). The following l integers indicate stamina of each person there.

The stamina of the person is a positive integer less that or equal to n + m.

Output

If it is possible to distribute tickets between people in the described manner print "YES", otherwise print "NO".

Examples

Input

2 2
3 3 3 2
1 3

Output

YES

Input

2 2
3 2 3 3
1 2

Output

NO

正解：贪心
解题报告：

　　因为我们想使得到两个出发点的距离小，但是不能同时保证两个，那么我们只能先保证一个最优，才想办法判断另外一个。显然把所有点按到左下角距离排序，可以对于左下角的点判断可达，然后我们每次走离左上角尽可能远的点，这样相当于是帮左上角的点分担了一部分远的点，同时可以保证合法性。

 //It is made by jump~
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>
 #include <ctime>
 #include <vector>
 #include <queue>
 #include <map>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int MAXN = ;
 int n,m,k,tot;
 int a[MAXN];
 struct node{
     int x,y,z,dis;
     bool operator < (const node &a) const{
     return a.z>z;
     }
 }s[MAXN];

 priority_queue<node>Q;
 inline int getint()
 {
        int w=,q=; char c=getchar();
        while((c<'' || c>'') && c!='-') c=getchar(); if(c=='-') q=,c=getchar();
        while (c>='' && c<='') w=w*+c-'', c=getchar(); return q ? -w : w;
 }

 inline bool cmp(node q,node qq){ return q.dis<qq.dis; }

 inline void work(){
     n=getint(); m=getint(); k=getint();
     for(int i=;i<=k;i++) a[i]=getint();
     sort(a+,a+k+);
     for(int i=;i<=n;i++)
     for(int j=;j<=m;j++)
         s[++tot].x=i,s[tot].y=j,s[tot].z=m+-j+i,s[tot].dis=i+j;
     sort(s+,s+tot+,cmp); int u=; bool ok=true;

     for(int i=;i<=k;i++) {
     while(a[i]>=s[u].dis && u<=tot) Q.push(s[u]),u++;
     if(Q.empty()) { ok=false; break;  }
     Q.pop();//清空
     }
     if(!ok) { printf("NO"); return; }
     while(u<=tot) Q.push(s[u]),u++;

     k=getint(); for(int i=;i<=k;i++)  a[i]=getint();
     sort(a+,a+k+);
     for(int i=k;i>=;i--) {
     if(a[i]<Q.top().z) { ok=false; break; }
     Q.pop();
     }
     if(!ok) { printf("NO"); return; }
     printf("YES");
 }

 int main()
 {
   work();
   return ;
 }