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;
typedef vector<int> vi;
const int inf=1<<30;

int h,w,q;
string s,t;
vi a,b,c[4];

int f(int x,int y,int X,int Y,bool fx=1,bool fy=1){
	int I=a[x],J=b[y];
	bool bx=0,by=0;
	if(I==1&&Y<=y||I==3&&y<=Y) bx=1;
	if(J==0&&X<=x||J==2&&x<=X) by=1;
	if(bx&&by){
		if(I==a[X]||J==b[Y]) return abs(x-X)+abs(y-Y);
		auto ix=lower_bound(c[I].begin(),c[I].end(),X);
		auto iy=lower_bound(c[J].begin(),c[J].end(),Y);
		if(J==0) ix--;
		if(I==1) iy--;
		int tx=(0<=*ix&&*ix<h?abs(*ix-x)+abs(*ix-X)+abs(y-Y):inf);
		int ty=(0<=*iy&&*iy<w?abs(*iy-y)+abs(*iy-Y)+abs(x-X):inf);
		return min(tx,ty);
	}
	auto ix=lower_bound(c[(I+2)%4].begin(),c[(I+2)%4].end(),x);
	auto iy=lower_bound(c[(J+2)%4].begin(),c[(J+2)%4].end(),y);
	if(J==0) ix--;
	if(I==1) iy--;
	int tx=(fx&&0<=*ix&&*ix<h?f(*ix,y,X,Y,0,1)+abs(*ix-x):inf);
	int ty=(fy&&0<=*iy&&*iy<w?f(x,*iy,X,Y,1,0)+abs(*iy-y):inf);
	return min(tx,ty);
}

int main(){
	cin>>h>>w>>q>>s>>t;
	a=vi(h,1);
	b=vi(w);
	for(int i=0;i<4;i++) c[i].push_back(-1);
	for(int i=0;i<h;i++){
		if(s[i]=='E') a[i]=3;
		c[a[i]].push_back(i);
	}
	for(int i=0;i<w;i++){
		if(t[i]=='S') b[i]=2;
		c[b[i]].push_back(i);
	}
	for(int i=0;i<4;i++) c[i].push_back(inf);
	for(int i=0;i<q;i++){
		int x,y,X,Y;
		cin>>x>>y>>X>>Y;
		int res=f(x-1,y-1,X-1,Y-1);
		cout<<(res>=inf?-1:res)<<endl;
	}
}