Problem: 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.

Examples

Input

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


Output

6
11
5
4


Input

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


Output

4
-1


Input

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


Output

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

Solution (in CPP):

Solution:
#include <stdio.h>
#include <algorithm>

#define N 0
#define S 1
#define E 2
#define W 3

using namespace std;

int n , m , q;
char ch[10] = "NSEW";
char s[120000] , t[120000];
int nxt[5][120000] , pre[5][120000];
int x1 , y1 , x2 , y2;
int ans;
int check ( int x , int y , int to , int del ) {
	int ok = 0;
	if ( (x==x2&&(to==E||to==W)) || (y==y2&&(to==N||to==S)) ) {
		ok = 1;
	}
	else {
		if ( to == N || to == S ) {
			if ( (y>y2&&s[x2]=='W') || (y<y2&&s[x2]=='E') ) ok = 1;
		}
		if ( to == E || to == W ) {
			if ( (x<x2&&t[y2]=='S') || (x>x2&&t[y2]=='N') ) ok = 1;
		}
	}
	//printf ( "chk %d %d %c %d %d\n" , x , y , ch[to] , del , ok );
	if ( !ok ) return 0;
	ans = min ( ans , del + abs(x-x2) + abs(y-y2) );
	return 1;
}
void dfs ( int dep , int x , int y , int to , int del ) {
	//printf ( "%d %d %d %c %d\n" , dep , x , y , ch[to] , del );
	if ( x == 0 || y == 0 ) return ;
	int face = 0;
	if ( to == N && x >= x2 ) face = 1;
	if ( to == S && x <= x2 ) face = 1;
	if ( to == E && y <= y2 ) face = 1;
	if ( to == W && y >= y2 ) face = 1;
	if ( face && check ( x , y , to , del ) ) return ;
	if ( dep > 4 ) return ;
	if ( to == N ) {
		dfs ( dep + 1 , pre[E][x] , y , E , del + abs(x-pre[E][x]) );
		dfs ( dep + 1 , pre[W][x] , y , W , del + abs(x-pre[W][x]) );
	}
	if ( to == S ) {
		dfs ( dep + 1 , nxt[E][x] , y , E , del + abs(x-nxt[E][x]) );
		dfs ( dep + 1 , nxt[W][x] , y , W , del + abs(x-nxt[W][x]) );
	}
	if ( to == E ) {
		dfs ( dep + 1 , x , nxt[N][y] , N , del + abs(y-nxt[N][y]) );
		dfs ( dep + 1 , x , nxt[S][y] , S , del + abs(y-nxt[S][y]) );
	}
	if ( to == W ) {
		dfs ( dep + 1 , x , pre[N][y] , N , del + abs(y-pre[N][y]) );
		dfs ( dep + 1 , x , pre[S][y] , S , del + abs(y-pre[S][y]) );
	}
	if ( face ) {
		if ( to == N ) {
			dfs ( dep + 1 , pre[E][x2] , y , E , del + abs(x-pre[E][x2]) );
			dfs ( dep + 1 , pre[W][x2] , y , W , del + abs(x-pre[W][x2]) );
		}
		if ( to == S ) {
			dfs ( dep + 1 , nxt[E][x2] , y , E , del + abs(x-nxt[E][x2]) );
			dfs ( dep + 1 , nxt[W][x2] , y , W , del + abs(x-nxt[W][x2]) );
		}
		if ( to == E ) {
			dfs ( dep + 1 , x , nxt[N][y2] , N , del + abs(y-nxt[N][y2]) );
			dfs ( dep + 1 , x , nxt[S][y2] , S , del + abs(y-nxt[S][y2]) );
		}
		if ( to == W ) {
			dfs ( dep + 1 , x , pre[N][y2] , N , del + abs(y-pre[N][y2]) );
			dfs ( dep + 1 , x , pre[S][y2] , S , del + abs(y-pre[S][y2]) );
		}
	}
}
void work () {
	int i , j;
	scanf ( "%d%d%d" , &n , &m , &q );
	scanf ( "%s" , s + 1 );
	scanf ( "%s" , t + 1 );
	for ( i = 1 ; i <= n ; i++ ) {
		pre[E][i] = pre[E][i-1];
		pre[W][i] = pre[W][i-1];
		if ( s[i] == 'E' ) pre[E][i] = i;
		else pre[W][i] = i;
	}
	for ( i = n ; i >= 1 ; i-- ) {
		nxt[E][i] = nxt[E][i+1];
		nxt[W][i] = nxt[W][i+1];
		if ( s[i] == 'E' ) nxt[E][i] = i;
		else nxt[W][i] = i;
	}
	for ( i = 1 ; i <= m ; i++ ) {
		pre[S][i] = pre[S][i-1];
		pre[N][i] = pre[N][i-1];
		if ( t[i] == 'S' ) pre[S][i] = i;
		else pre[N][i] = i;
	}
	for ( i = m ; i >= 1 ; i-- ) {
		nxt[S][i] = nxt[S][i+1];
		nxt[N][i] = nxt[N][i+1];
		if ( t[i] == 'S' ) nxt[S][i] = i;
		else nxt[N][i] = i;
	}
	//for ( i = 0 ; i <= 3 ; i++ )
	//	for ( j = 1 ; j <= m ; j++ )
	//		printf ( "%d %d %d %d\n" , i , j , pre[i][j] , nxt[i][j] );
	for ( i = 1 ; i <= q ; i++ ) {
		scanf ( "%d%d%d%d" , &x1 , &y1 , &x2 , &y2 );
		ans = 21000000;
		dfs ( 0 , x1 , y1 , s[x1]=='E'?E:W , 0 );
		dfs ( 0 , x1 , y1 , t[y1]=='S'?S:N , 0 );
		if ( ans == 21000000 ) ans = -1;
		printf ( "%d\n" , ans );
	}
}
int main () {
	work ();
	return 0;
}