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 ll long long
#define INF 1e9
#define maxn 100010
using namespace std;
template <typename T> void read(T &x){
	x=0;char c=getchar();int fh=1;
	while (!isdigit(c)){if (c=='-')fh=-1;c=getchar();}
	while (isdigit(c))x=x*10+c-'0',c=getchar();
	x*=fh;
}
struct Info{int x,y,di;}t;
int m,n,q,xl,yl,xr,yr,a[20],b[20],na,nb,nc,c[20],di[12][12];
string s1,s2;
set<int> W,E,N,S;
bool operator <(Info x,Info y){return x.di<y.di;}
int main(){
	read(n);read(m);read(q);
	cin>>s1;cin>>s2;
	for (int i=0;i<n;i++) if (s1[i]=='W')W.insert(i+1);else E.insert(i+1);
	for (int i=0;i<m;i++) if (s2[i]=='N')N.insert(i+1);else S.insert(i+1);
	while (q--){
		read(xl);read(yl);read(xr);read(yr);
		memset(a,0,sizeof(a));memset(b,0,sizeof(b));na=nb=0;
		set<int>::iterator p;
		p=W.lower_bound(xl);
		if (p!=W.end())a[++na]=*p;
		if (p!=W.begin()&&(*p!=xl)){p--;a[++na]=*p;}
		p=W.lower_bound(xr);
		if (p!=W.end())a[++na]=*p;
		if (p!=W.begin()&&(*p!=xr)){p--;a[++na]=*p;}
		p=E.lower_bound(xl);
		if (p!=E.end())a[++na]=*p;
		if (p!=E.begin()&&(*p!=xl)){p--;a[++na]=*p;}
		p=E.lower_bound(xr);
		if (p!=E.end())a[++na]=*p;
		if (p!=E.begin()&&(*p!=xr)){p--;a[++na]=*p;}
		p=S.lower_bound(yl);
		if (p!=S.end())b[++nb]=*p;
		if (p!=S.begin()&&(*p!=yl)){p--;b[++nb]=*p;}
		p=S.lower_bound(yr);
		if (p!=S.end())b[++nb]=*p;
		if (p!=S.begin()&&(*p!=yr)){p--;b[++nb]=*p;}
		p=N.lower_bound(yl);
		if (p!=N.end())b[++nb]=*p;
		if (p!=N.begin()&&(*p!=yl)){p--;b[++nb]=*p;}
		p=N.lower_bound(yr);
		if (p!=N.end())b[++nb]=*p;
		if (p!=N.begin()&&(*p!=yr)){p--;b[++nb]=*p;}
		sort(a+1,a+na+1);
		sort(b+1,b+nb+1);
		for (int i=1;i<=na;i++)c[i]=a[i];
		nc=na;na=0;
		for (int i=1;i<=nc;i++)if (i==1||c[i]!=c[i-1]) a[++na]=c[i];
		for (int i=1;i<=nb;i++)c[i]=b[i];
		nc=nb;nb=0;
		for (int i=1;i<=nc;i++)if (i==1||c[i]!=c[i-1]) b[++nb]=c[i];
		for (int i=1;i<=na;i++)
			for (int j=1;j<=nb;j++)di[i][j]=INF;
		priority_queue<Info> s;while (!s.empty()) s.pop();
		for (int i=1;i<=na;i++)
			for (int j=1;j<=nb;j++)
				if (a[i]==xl&&b[j]==yl){di[i][j]=0;s.push((Info){i,j,0});}
//		cout<<na<<' '<<nb<<endl;
//		for (int i=1;i<=na;i++) cout<<a[i]<<' ';cout<<endl;
//		for (int i=1;i<=nb;i++) cout<<b[i]<<' ';cout<<endl;
		while (!s.empty()){
			t=s.top();s.pop();
//			cout<<t.x<<' '<<t.y<<' '<<t.di<<endl;
			int nx=t.x,ny=t.y;
			if (s2[b[ny]-1]=='N')nx--;else nx++;
			if (nx>0&&nx<=na){
				if (di[nx][ny]>t.di+abs(a[nx]-a[t.x])){
					di[nx][ny]=t.di+abs(a[nx]-a[t.x]);
					s.push((Info){nx,ny,di[nx][ny]});
				}
			}
			nx=t.x,ny=t.y;
			if (s1[a[nx]-1]=='W')ny--;else ny++;
			if (ny>0&&ny<=nb){
				if (di[nx][ny]>t.di+abs(b[ny]-b[t.y])){
					di[nx][ny]=t.di+abs(b[ny]-b[t.y]);
					s.push((Info){nx,ny,di[nx][ny]});
				}
			}
		}
		for (int i=1;i<=na;i++)
			for (int j=1;j<=nb;j++)
				if (a[i]==xr&&b[j]==yr){if (di[i][j]==INF) puts("-1");else printf("%d\n",di[i][j]);}
	}
	return 0;
}