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;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())

int n,m,q;
char s[100005];
char t[100005];
vector<int>up,dw,le,ri;
ll ans;
void  on_y(int a,int b,int c,int d,int dep,ll cur_ans);

void  on_x(int a,int b,int c,int d,int dep,ll cur_ans){
	if(dep == 0) return ;
	if(s[a] == 'W'){
		//左
		int x = POSL(up,min(d,b));
		if(x){
			x--;
			on_y(a,up[x],c,d,dep-1,cur_ans+abs(b-up[x]));
		}
		int y = POSL(dw,min(d,b));
		if(y){
			y--;
			on_y(a,dw[y],c,d,dep-1,cur_ans+abs(b-dw[y]));
		}
		if(d <= b){
			if(t[d] == 'N' && c <= a){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
			if(t[d] == 'S' && c >= a){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
		}
	}
	else{
		//右
		int x = POSU(up,max(d,b));
		if(x < up.size()){
			on_y(a,up[x],c,d,dep-1,cur_ans+abs(b-up[x]));
		}
		int y = POSU(dw,max(d,b));
		if(y < dw.size()){
			on_y(a,dw[y],c,d,dep-1,cur_ans+abs(b-dw[y]));
		}
		if(d >= b){
			if(t[d] == 'N' && c <= a){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
			if(t[d] == 'S' && c >= a){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
		}
	}
	return ;
}
void  on_y(int a,int b,int c,int d,int dep,ll cur_ans){
	if(dep == 0) return ;
	if(t[b] == 'N'){
		//上
		int x = POSL(le,min(a,c));
		if(x){
			x--;
			on_x(le[x],b,c,d,dep-1,cur_ans+abs(a-le[x]));
		}
		int y = POSL(ri,min(a,c));
		if(y){
			y--;
			on_x(ri[y],b,c,d,dep-1,cur_ans+abs(a-ri[y]));
		}
		if(c <= a){
			if(s[c] == 'W' && d <= b){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
			if(s[c] == 'E' && d >= b){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
		}
	}
	else{
		//右
		int x = POSU(le,max(a,c));
		if(x < le.size()){
			on_x(le[x],b,c,d,dep-1,cur_ans+abs(a-le[x]));
		}
		int y = POSU(ri,max(a,c));
		if(y < ri.size()){
			on_x(ri[y],b,c,d,dep-1,cur_ans+abs(a-ri[y]));
		}
		if(c >= a){
			if(s[c] == 'W' && d <= b){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
			if(s[c] == 'E' && d >= b){
				ans = min(ans,cur_ans+abs(a-c)+abs(b-d));
			}
		}
	}
	return ;
}
int main(){
	scanf("%d%d%d",&n,&m,&q);
	scanf("%s",&s);
	for(int i=0;i<n;i++){
		if(s[i] == 'W') le.pb(i);
		else ri.pb(i);
	}
	scanf("%s",&t);
	for(int i=0;i<m;i++){
		if(t[i] == 'N') up.pb(i);
		else dw.pb(i);
	}
	for(int i=0;i<q;i++){
		int a,b,c,d; scanf("%d%d%d%d",&a,&b,&c,&d);
		a--; b--; c--; d--;
		ans = 1e18;
		on_x(a,b,c,d,7,0);
		on_y(a,b,c,d,7,0);
		if(ans > 1e17) puts("-1");
		else printf("%lld\n",ans);
	}
}