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: Accepted)

#include <bits/stdc++.h>
using namespace std;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while ('0'>ch||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do {(t*=10)+=ch-'0';ch=getchar();} while ('0'<=ch&&ch<='9'); t*=f;
}
const int maxn=(3e5)+10;
int n,w,h,ans,mx[maxn*4],lazy[maxn*4];
int a[maxn],b[maxn],t1,t2;
struct node { int x,y; } d[maxn];
bool cmp(node a,node b) {
	return a.x<b.x;
}
void pushdown(int root) {
	if (lazy[root]==0) return;
	mx[root*2]+=lazy[root];
	mx[root*2+1]+=lazy[root];
	lazy[root*2]+=lazy[root];
	lazy[root*2+1]+=lazy[root];
	lazy[root]=0;
}
void add(int L,int R,int l,int r,int root,int delta) {
	if (L<=l&&r<=R) {
		mx[root]+=delta; lazy[root]+=delta; return;
	} pushdown(root);
	int mid=(l+r)/2;
	if (L<=mid) add(L,R,l,mid,root*2,delta);
	if (mid<R) add(L,R,mid+1,r,root*2+1,delta);
	mx[root]=max(mx[root*2],mx[root*2+1]);
}
void solve() {
	sort(d+1,d+n+1,cmp);
	memset(mx,0,sizeof(mx));
	memset(lazy,0,sizeof(lazy));
	t1=t2=a[1]=b[1]=1; add(1,n,1,n,1,h);
	for (int i=2;i<=n;i++) {
		add(1,i-1,1,n,1,d[i].x-d[i-1].x);
		ans=max(ans,mx[1]);
		if (d[i].y>=h/2) {
			add(a[t1],i-1,1,n,1,d[i].y-h);
			while (t1>1&&d[a[t1]].y>=d[i].y)
				add(a[t1-1],a[t1]-1,1,n,1,d[i].y-d[a[t1]].y),t1--;
			a[++t1]=i;
		} else {
			add(b[t2],i-1,1,n,1,-d[i].y);
			while (t2>1&&d[b[t2]].y<=d[i].y)
				add(b[t2-1],b[t2]-1,1,n,1,d[b[t2]].y-d[i].y),t2--;
			b[++t2]=i;
		}
	}
}
int main() {
	read(w); read(h); read(n);
	for (int i=1;i<=n;i++) read(d[i].x),read(d[i].y);
	d[++n]=(node){0,0}; d[++n]=(node){w,h};
	solve(); swap(w,h); for (int i=1;i<=n;i++) swap(d[i].x,d[i].y);
	solve(); printf("%d\n",ans*2);
	return 0;
}