Problem:
Score : 1200 points

Problem Statement

In Takaha-shi, the capital of Republic of AtCoder, there are N roads extending east and west, and M roads extending north and south. There are no other roads.
The i-th east-west road from the north and the j-th north-south road from the west cross at the intersection (i, j).
Two east-west roads do not cross, nor do two north-south roads.
The distance between two adjacent roads in the same direction is 1.

Each road is one-way; one can only walk in one direction. The permitted direction for each road is described by a string S of length N and a string T of length M, as follows:

If the i-th character in S is W, one can only walk westward along the i-th east-west road from the north;

If the i-th character in S is E, one can only walk eastward along the i-th east-west road from the north;

If the i-th character in T is N, one can only walk northward along the i-th north-south road from the west;

If the i-th character in T is S, one can only walk southward along the i-th south-west road from the west.

Process the following Q queries:

In the i-th query, a_i, b_i, c_i and d_i are given. What is the minimum distance to travel to reach the intersection (c_i, d_i) from the intersection (a_i, b_i) by walking along the roads?

Constraints

2 \leq N \leq 100000

2 \leq M \leq 100000

2 \leq Q \leq 200000

|S| = N

S consists of W and E.

|T| = M

T consists of N and S.

1 \leq a_i \leq N

1 \leq b_i \leq M

1 \leq c_i \leq N

1 \leq d_i \leq M

(a_i, b_i) \neq (c_i, d_i)

Input

Input is given from Standard Input in the following format:

N M Q
S
T
a_1 b_1 c_1 d_1
a_2 b_2 c_2 d_2
:
a_Q b_Q c_Q d_Q

Output

In the i-th line, print the response to the i-th query. If the intersection (c_i, d_i) cannot be reached from the intersection (a_i, b_i) by walking along the roads, print -1 instead.

Sample Input 1

4 5 4
EEWW
NSNNS
4 1 1 4
1 3 1 2
4 2 3 2
3 3 3 5

Sample Output 1

6
11
5
4

The permitted direction for each road is shown in the following figure (north upward):

For each of the four queries, a route that achieves the minimum travel distance is as follows:

Sample Input 2

3 3 2
EEE
SSS
1 1 3 3
3 3 1 1

Sample Output 2

4
-1

The travel may be impossible.

Sample Input 3

9 7 10
EEEEEWEWW
NSSSNSN
4 6 9 2
3 7 6 7
7 5 3 5
1 1 8 1
4 3 5 4
7 4 6 4
2 5 8 6
6 6 2 7
2 4 7 5
7 2 9 7

Sample Output 3

9
-1
4
9
2
3
7
7
6
-1

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

#include<bits/stdc++.h>
using namespace std;
#define nn 200008
#define pii pair<int,int>
#define mp make_pair
int n,m;int q;pii from[nn],dest[nn];
int ans[nn];
int dir[2][nn];
vector<int> path[2][2];
/*
0...
....
...1

0->
1|
*/
int lb(int x,vector<int> &v) {if(v[0]>x) return 0;return v[upper_bound(v.begin(),v.end(),x)-v.begin()-1];}
int ub(int x,vector<int> &v) {if(v[v.size()-1]<x) return 0;return v[lower_bound(v.begin(),v.end(),x)-v.begin()];}


void solve()
{
	path[0][0].clear(),path[0][1].clear(),path[1][0].clear(),path[1][1].clear();
	
	for(int i=1;i<=n;i++)
		path[0][dir[0][i]].push_back(i);
	
	
	for(int i=1;i<=m;i++)
		path[1][dir[1][i]].push_back(i);

	for(int i=1;i<=q;i++)	
	{
		pii s=from[i],t=dest[i];
		
		if(s.first>t.first or s.second>t.second) continue;
		#define case0 case0
		for(;s.second==t.second;)
		{
			int x=ub(s.second,path[1][1]);
			if(x==s.second) {ans[i]=abs(s.first-t.first);break;}
			int y=lb(s.first,path[0][1]),z=ub(t.first,path[0][0]);
			if(x&&y&&z) ans[i]=min(ans[i],abs(s.first-t.first)+abs(y-s.first)*2+abs(s.second-x)*2+abs(t.first-z)*2);
			break;
		} 
		if(s.second==t.second) continue;
		#define case1 case1
		for(;s.first<=t.first;)
		{
			int x=lb(s.first,path[0][1]);
			if(!x) break;int tmp=abs(s.first-x)+abs(t.first-x)+abs(s.second-t.second);
			if(x!=s.first) 
			{
				int y=lb(s.second,path[1][0]);if(!y) break;
				tmp+=abs(y-s.second)*2;
			}
			int otmp=tmp;

			for(;1;)
			{
				int y=ub(t.second,path[1][1]);if(!y) break;
				tmp+=abs(y-t.second)*2;
				if(y!=t.second)
				{
					int z=ub(t.first,path[0][0]);if(!z) break;
					tmp+=abs(z-t.first)*2;
				}
				ans[i]=min(ans[i],tmp);
				break;
			}
			tmp=otmp;
			for(;0;)
			{
				int y=lb(t.second,path[1][1]);if(!y) break;
				if(y!=t.second)
				{
					int z=ub(t.first,path[0][1]);if(!z) break;
					tmp+=abs(z-t.first)*2;
				}
				ans[i]=min(ans[i],tmp);
				break;
			}
			break;
		}
		#define case2 case2
		for(;s.first<t.first;)//middle
		{
			int x=ub(s.first,path[0][1]);
			if(x>=t.first or x<=s.first) break;int tmp=abs(s.second-t.second)+abs(s.first-t.first);
			int at=1e9,bt=1e9;int otmp=tmp;
			for(;1;)
			{
				int y=ub(s.second,path[1][1]);if(!y) break;
				tmp+=abs(y-s.second)*2;
				if(y!=s.second)
				{
					int z=lb(s.second,path[0][1]);
					if(!z) break;tmp+=abs(z-s.first)*2;
				}	
				at=tmp;
				break;
			}
			tmp=otmp;
			for(;1;)
			{
				int y=lb(s.second,path[1][1]);if(!y) break;
				tmp+=abs(y-s.second)*2;
				bt=tmp;
				break;
			}
			tmp=min(at,bt);
			
			//copied from  line58
			otmp=tmp;

			for(;1;)
			{
				int y=ub(t.second,path[1][1]);if(!y) break;
				tmp+=abs(y-t.second)*2;
				if(y!=t.second)
				{
					int z=ub(t.first,path[0][0]);if(!z) break;
					tmp+=abs(z-t.first)*2;
				}
				ans[i]=min(ans[i],tmp);
				break;
			}
			tmp=otmp;
			for(;0;)
			{
				int y=lb(t.second,path[1][1]);if(!y) break;
				if(y!=t.second)
				{
					int z=ub(t.first,path[0][1]);if(!z) break;
					tmp+=abs(z-t.first)*2;
				}
				ans[i]=min(ans[i],tmp);
				break;
			}
			break;
			//licensed
		}
	}
}


void transform(int mask)
{
	for(int i=1;i<=n;i++) if(mask&1) dir[0][i]^=1;
	if(mask&2) reverse(dir[0]+1,dir[0]+n+1);
	for(int i=1;i<=m;i++) if(mask&2) dir[1][i]^=1;
	if(mask&1) reverse(dir[1]+1,dir[1]+m+1);
	
	for(int i=1;i<=q;i++)
	{
		if(mask&2) 
		{
			from[i].first=n-from[i].first+1;
			dest[i].first=n-dest[i].first+1;
		}
		if(mask&1)
		{
			from[i].second=m-from[i].second+1;
			dest[i].second=m-dest[i].second+1;
		}
	}
}
char buf[nn];
int main()
{
	scanf("%d%d%d",&n,&m,&q);
	scanf("%s",buf+1);for(int i=1;i<=n;i++) dir[0][i]=(buf[i]=='E');
	scanf("%s",buf+1);for(int i=1;i<=m;i++) dir[1][i]=(buf[i]=='S');
	
	
	
	for(int i=1;i<=q;i++) ans[i]=1e9,scanf("%d%d%d%d",&from[i].first,&from[i].second,&dest[i].first,&dest[i].second);	
	
	for(int i=0;i<4;i++) {transform(i);solve();transform(i);}	
	
	
	
	for(int i=1;i<=n;i++) dir[0][i]^=1;for(int i=1;i<=m;i++) dir[1][i]^=1;
	for(int i=1;i<=q;i++) swap(from[i],dest[i]);
	
	
	
	for(int i=0;i<4;i++) {transform(i);solve();transform(i);}	
	
	for(int i=1;i<=q;i++) if(ans[i]<1e7)printf("%d\n",ans[i]);else printf("%d\n",-1);
	return 0;
}