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 <bits/stdc++.h>
using namespace std ;
template < class R > inline void Readin( R &K ) {
	K = 0 ; register bool F = false ; register char C = getchar() ;
	while( C < '0' or C > '9' ) { if( C == '-' ) F = true ; C = getchar() ; }
	while( C <= '9' and C >= '0' ) K = ( K << 1 ) + ( K << 3 ) + C - '0' , C = getchar() ;
	if( F ) K = -K ;
}
const int MaxN = 100000 + 10 ;
int N , M , Q ;
bool Tr[MaxN] , Td[MaxN] ;
int Upu[MaxN] , Upd[MaxN] , Upl[MaxN] , Upr[MaxN] ;
int Lou[MaxN] , Lod[MaxN] , Lol[MaxN] , Lor[MaxN] ;
char S[MaxN] ;
int X , Y , Ans ;
int Ls , Rs , Us , Ds , C ;
bool Tol , Tor , Tou , Tod , Best ;
void Dfs( int A , int B , int Tans , int Step , short Last ) {
	if( A == X and B == Y ) {
		Ans = min( Ans , Tans ) ;
		if( Ans == abs( A - X ) + abs( B - Y ) ) Best = true ;
		return ;
	}
	if( Ls == 3 or Rs == 3 or Us == 3 or Ds == 3 or C == 3 or (++Step) == 8 or Tans >= Ans or Best )
		return ;
	register int To1 , To2 ;
	if( Last != 0 ) {
		if( Tr[A] ) {
			++Rs ;
			To1 = Upu[B+1] ;
			if( To1 != -1 ) Dfs( A , To1 , Tans + To1 - B , Step , 0 ) ;
			if( Y > B ) {
				++C ;
				To2 = Upu[Y] ;
				if( To2 != -1 and To2 != To1 ) Dfs( A , To2 , Tans + To2 - B , Step , 0 ) ;
				--C ;
			}
			To1 = Upd[B+1] ;
			if( To1 != -1 ) Dfs( A , To1 , Tans + To1 - B , Step , 0 ) ;
			if( Y > B ) {
				++C ;
				To2 = Upd[Y] ;
				if( To2 != -1 and To2 != To1 ) Dfs( A , To2 , Tans + To2 - B , Step , 0 ) ;
				--C ;
			}
			--Rs ;
		}
		else {
			++Ls ;
			To1 = Lou[B-1] ;
			if( To1 != -1 ) Dfs( A , To1 , Tans - To1 + B , Step , 0 ) ;
			if( Y < B ) {
				++C ;
				To2 = Lou[Y] ;
				if( To2 != -1 and To2 != To1 ) Dfs( A , To2 , Tans - To2 + B , Step , 0 ) ;
				--C ;
			}
			To1 = Lod[B-1] ;
			if( To1 != -1 ) Dfs( A , To1 , Tans - To1 + B , Step , 0 ) ;
			if( Y < B ) {
				++C ;
				To2 = Lod[Y] ;
				if( To2 != -1 and To2 != To1 ) Dfs( A , To2 , Tans - To2 + B , Step , 0 ) ;
				--C ;
			}
			--Ls ;
		}
	}
	if( Last != 1 ) {
		if( Td[B] ) {
			++Us ;
			To1 = Upl[A+1] ;
			if( To1 != -1 ) Dfs( To1 , B , Tans + To1 - A , Step , 1 ) ;
			if( X > A ) {
				++C ;
				To2 = Upl[X] ;
				if( To2 != -1 and To2 != To1 ) Dfs( To2 , B , Tans + To2 - A , Step , 1 ) ;
				--C ;
			}
			To1 = Upr[A+1] ;
			if( To1 != -1 ) Dfs( To1 , B , Tans + To1 - A , Step , 1 ) ;
			if( X > A ) {
				++C ;
				To2 = Upr[X] ;
				if( To2 != -1 and To2 != To1 ) Dfs( To2 , B , Tans + To2 - A , Step , 1 ) ;
				--C ;
			}
			--Us ;
		}
		else {
			++Ds ;
			To1 = Lol[A-1] ;
			if( To1 != -1 ) Dfs( To1 , B , Tans - To1 + A , Step , 1 ) ;
			if( X < A ) {
				++C ;
				To2 = Lol[X] ;
				if( To2 != -1 and To2 != To1 ) Dfs( To2 , B , Tans - To2 + A , Step , 1 ) ;
				--C ;
			}
			To1 = Lor[A-1] ;
			if( To1 != -1 ) Dfs( To1 , B , Tans - To1 + A , Step , 1 ) ;
			if( X < A ) {
				++C ;
				To2 = Lor[X] ;
				if( To2 != -1 and To2 != To1 ) Dfs( To2 , B , Tans - To2 + A , Step , 1 ) ;
				--C ;
			}
			--Ds ;
		}
	}
}
int main() {
	Readin( N ) ;
	Readin( M ) ;
	Readin( Q ) ;
	scanf( "%s" , S + 1 ) ; getchar() ;
	for(register int i = 0 ; ++i <= N ; )
		Tr[i] = ( S[i] == 'E' ) ;
	scanf( "%s" , S + 1 ) ; getchar() ;
	for(register int i = 0 ; ++i <= M ; )
		Td[i] = ( S[i] == 'S' ) ;
	Upl[N+1] = Upr[N+1] = -1 ;
	for(register int i = N ; i ; --i ) {
		Upl[i] = Upl[i+1] ;
		Upr[i] = Upr[i+1] ;
		Tr[i] ? Upr[i] = i : Upl[i] = i ;
	}
	Lol[0] = Lor[0] = -1 ;
	for(register int i = 0 ; ++i <= N ; ) {
		Lol[i] = Lol[i-1] ;
		Lor[i] = Lor[i-1] ;
		Tr[i] ? Lor[i] = i : Lol[i] = i ;
	}
	Upu[M+1] = Upd[M+1] = -1 ;
	for(register int i = M ; i ; --i ) {
		Upu[i] = Upu[i+1] ;
		Upd[i] = Upd[i+1] ;
		Td[i] ? Upd[i] = i : Upu[i] = i ;
	}
	Lou[0] = Lod[0] = -1 ;
	for(register int i = 0 ; ++i <= M ; ) {
		Lou[i] = Lou[i-1] ;
		Lod[i] = Lod[i-1] ;
		Td[i] ? Lod[i] = i : Lou[i] = i ;
	}
	while( Q-- ) {
		register int A , B ;
		Ans = 1 << 30 ;
		Best = false ;
		Readin( A ) ;
		Readin( B ) ;
		Readin( X ) ;
		Readin( Y ) ;
		Tol = Y <= B ;
		Tor = Y >= B ;
		Tou = X <= A ;
		Tod = X >= A ;
		Dfs( A , B , 0 , 0 , 233 ) ;
		if( Ans == ( 1 << 30 ) ) printf( "-1\n" ) ;
			else printf( "%d\n" , Ans ) ;
	}
	return 0 ;
}