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>
#define fi first
#define se second
#define pb push_back
#define SZ(x) ((int)x.size())
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=(a); i<=(b); i++)
#define per(i,a,b) for (register int i=(a); i>=(b); i--)
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef pair<int,int> Pii;
typedef vector<int> Vi;
template<class T> inline void read(T &x){
	x=0; char c=getchar(); int f=1;
	while (!isdigit(c)) {if (c=='-') f=-1; c=getchar();}
	while (isdigit(c)) {x=x*10+c-'0'; c=getchar();} x*=f;
}
template<class T> inline void umin(T &x, T y){x=x<y?x:y;}
template<class T> inline void umax(T &x, T y){x=x>y?x:y;}
inline ui R() {
	static ui seed=416;
	return seed^=seed>>5,seed^=seed<<17,seed^=seed>>13;
}
const int N = 266666,inf=0x3f3f3f3f;
int n,m,q,res;char S[N],T[N];
int qz1[N][2],hz1[N][2],hz0[N][2],qz0[N][2];//0:- 1:+
int Abs(int x){return x>0?x:-x;}
void dfs(int x1, int y1, int x2, int y2, int cur, bool dir, int dep){
	if(dep>4)return;
//	printf("dfs %d %d %d %d\n",x1,y1,dir,cur);
	if(!dir){
		if(S[x1]=='E'){
			if(x1==x2&&y1<y2){umin(res,cur+Abs(x2-x1)+Abs(y2-y1));return;}
			if(hz1[y1][0])dfs(x1,hz1[y1][0],x2,y2,cur+Abs(hz1[y1][0]-y1),dir^1,dep+1);
			if(hz1[y1][1])dfs(x1,hz1[y1][1],x2,y2,cur+Abs(hz1[y1][1]-y1),dir^1,dep+1);
			if(y2>y1&&hz1[y2][x2>x1])dfs(x1,hz1[y2][x2>x1],x2,y2,cur+Abs(hz1[y2][x2>x1]-y1),dir^1,dep+1);
		}
		else{
			if(x1==x2&&y1>y2){umin(res,cur+Abs(x2-x1)+Abs(y2-y1));return;}
			if(qz1[y1][0])dfs(x1,qz1[y1][0],x2,y2,cur+Abs(qz1[y1][0]-y1),dir^1,dep+1);
			if(qz1[y1][1])dfs(x1,qz1[y1][1],x2,y2,cur+Abs(qz1[y1][1]-y1),dir^1,dep+1);
			if(y2<y1&&qz1[y2][x2>x1])dfs(x1,qz1[y2][x2>x1],x2,y2,cur+Abs(qz1[y2][x2>x1]-y1),dir^1,dep+1);
		}
	}
	else{
		if(T[y1]=='S'){
			if(y1==y2&&x1<x2){umin(res,cur+Abs(x2-x1)+Abs(y2-y1));return;}
			if(hz0[x1][0])dfs(hz0[x1][0],y1,x2,y2,cur+Abs(hz0[x1][0]-x1),dir^1,dep+1);
			if(hz0[x1][1])dfs(hz0[x1][1],y1,x2,y2,cur+Abs(hz0[x1][1]-x1),dir^1,dep+1);
			if(x2>x1&&hz0[x2][y2>y1])dfs(hz0[x2][y2>y1],y1,x2,y2,cur+Abs(hz0[x2][y2>y1]-x1),dir^1,dep+1);
		}
		else{
			if(y1==y2&&x1>x2){umin(res,cur+Abs(x2-x1)+Abs(y2-y1));return;}
			if(qz0[x1][0])dfs(qz0[x1][0],y1,x2,y2,cur+Abs(qz0[x1][0]-x1),dir^1,dep+1);
			if(qz0[x1][1])dfs(qz0[x1][1],y1,x2,y2,cur+Abs(qz0[x1][1]-x1),dir^1,dep+1);
			if(x2<x1&&qz0[x2][y2>y1])dfs(qz0[x2][y2>y1],y1,x2,y2,cur+Abs(qz0[x2][y2>y1]-x1),dir^1,dep+1);
		}
	}
}
int main() {
	read(n);read(m);read(q);
	scanf("%s%s",S+1,T+1);
	per(i,m,1){
		memcpy(hz1[i],hz1[i+1],sizeof(hz1[i+1]));
		if(T[i]=='N')hz1[i][0]=i;else hz1[i][1]=i;
	}
	rep(i,1,m){
		memcpy(qz1[i],qz1[i-1],sizeof(qz1[i-1]));
		if(T[i]=='N')qz1[i][0]=i;else qz1[i][1]=i;
	}
	per(i,n,1){
		memcpy(hz0[i],hz0[i+1],sizeof(hz0[i+1]));
		if(S[i]=='W')hz0[i][0]=i;else hz0[i][1]=i;
	}
	rep(i,1,n){
		memcpy(qz0[i],qz0[i-1],sizeof(qz0[i-1]));
		if(S[i]=='W')qz0[i][0]=i;else qz0[i][1]=i;
	}
	while(q--){
		int x1,y1,x2,y2;
		read(x1);read(y1);read(x2);read(y2);
		res=inf;dfs(x1,y1,x2,y2,0,0,0);dfs(x1,y1,x2,y2,0,1,0);
		printf("%d\n",res<inf?res:-1);
	}
	return 0;
}