Problem:
Score : 1600 points

Problem Statement

There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white.

Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i).

Then, for each 1 ≦ i ≦ N, he will paint one of the following four regions black:

the region satisfying x < x_i within the rectangle

the region satisfying x > x_i within the rectangle

the region satisfying y < y_i within the rectangle

the region satisfying y > y_i within the rectangle

Find the longest possible perimeter of the white region of a rectangular shape within the rectangle after he finishes painting.

Constraints

1 ≦ W, H ≦ 10^8

1 ≦ N ≦ 3 \times 10^5

0 ≦ x_i ≦ W (1 ≦ i ≦ N)

0 ≦ y_i ≦ H (1 ≦ i ≦ N)

W, H (21:32, added), x_i and y_i are integers.

If i ≠ j, then x_i ≠ x_j and y_i ≠ y_j.

Input

The input is given from Standard Input in the following format:

W H N
x_1 y_1
x_2 y_2
:
x_N y_N

Output

Print the longest possible perimeter of the white region of a rectangular shape within the rectangle after Snuke finishes painting.

Sample Input 1

10 10 4
1 6
4 1
6 9
9 4

Sample Output 1

32

In this case, the maximum perimeter of 32 can be obtained by painting the rectangle as follows:

Sample Input 2

5 4 5
0 0
1 1
2 2
4 3
5 4

Sample Output 2

12

Sample Input 3

100 100 8
19 33
8 10
52 18
94 2
81 36
88 95
67 83
20 71

Sample Output 3

270

Sample Input 4

100000000 100000000 1
3 4

Sample Output 4

399999994

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define x first
#define y second
using namespace std;
int w,h,n,ans,st1[300010],st2[300010];
pair<int,int> dt[300010];
struct seg
{
	int mx[3000100],tag[3000100];
	void clear()
	{
		memset(mx,0,sizeof(mx));
		memset(tag,0,sizeof(tag));
	}
	void pushdown(int root)
	{
		mx[2*root]+=tag[root];tag[2*root]+=tag[root];
		mx[2*root+1]+=tag[root];tag[2*root+1]+=tag[root];
		tag[root]=0;
	}
	void modify(int ql,int qr,int w,int root=1,int pl=1,int pr=n)
	{
		if ((pl==ql)&&(pr==qr)) {tag[root]+=w;mx[root]+=w;return;}
		pushdown(root);
		int mid=(pl+pr)>>1;
		if (ql<=mid) modify(ql,min(mid,qr),w,root*2,pl,mid);
		if (qr>mid) modify(max(mid+1,ql),qr,w,root*2+1,mid+1,pr);
		mx[root]=max(mx[root*2],mx[root*2+1]);
	}
}seg;
void work()
{
	sort(dt+1,dt+n+1);
	seg.clear();int top1=0,top2=0;
	if (dt[n].x!=w) dt[++n]=make_pair(w,-1);
	for (int i=1;i<=n;i++) seg.modify(i,i,-dt[i].x);
	for (int i=1;i<=n;i++)
	{
		seg.modify(i,i,h);
		ans=max(ans,dt[i].x+seg.mx[1]);
		if (dt[i].y>h/2)
		{
			seg.modify(st1[top1]+1,i,dt[i].y-h);
			while ((top1)&&(dt[st1[top1]].y>=dt[i].y))
			{
				seg.modify(st1[top1-1]+1,st1[top1],dt[i].y-dt[st1[top1]].y);
				top1--;
			}
			st1[++top1]=i;
		}
		else
		{
			seg.modify(st2[top2]+1,i,-dt[i].y);
			while ((top2)&&(dt[st2[top2]].y<=dt[i].y))
			{
				seg.modify(st2[top2-1]+1,st2[top2],dt[st2[top2]].y-dt[i].y);
				top2--;
			}
			st2[++top2]=i;
		}
	}
	n-=(dt[n].y<0);
}
int main()
{
	scanf("%d%d%d",&w,&h,&n);
	for (int i=1;i<=n;i++) scanf("%d%d",&dt[i].x,&dt[i].y);
	work();
	swap(w,h);
	for (int i=1;i<=n;i++) swap(dt[i].x,dt[i].y);
	work();
	printf("%d\n",ans*2);
}