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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
inline int read()
{
	int x=0;bool t=false;char ch=getchar();
	while((ch<'0'||ch>'9')&&ch!='-')ch=getchar();
	if(ch=='-')t=true,ch=getchar();
	while(ch<='9'&&ch>='0')x=x*10+ch-48,ch=getchar();
	return t?-x:x;
}
vector<int> N,W,E,S;
int Findpre(vector<int> &A,int x)
{
	if(!A.size())return 1;
	int l=0,r=A.size()-1,ret=0;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(A[mid]<=x)ret=mid,l=mid+1;
		else r=mid-1;
	}
	return A[ret];
}
int Findsuf(vector<int> &A,int x)
{
	if(!A.size())return 1;
	int l=0,r=A.size()-1,ret=r;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(A[mid]>=x)ret=mid,r=mid-1;
		else l=mid+1;
	}
	return A[ret];
}
int n,m,Q;char dS[100100],dT[100100];
struct Line{int v,next,w;}e[1000];
int bh[20][20],tot;
int h[100],cnt;
inline void Add(int u,int v,int w){e[cnt]=(Line){v,h[u],w};h[u]=cnt++;}
int dis[200];bool vis[200];
int SPFA(int S,int T)
{
	for(int i=1;i<=tot;++i)dis[i]=1e9,vis[i]=false;
	queue<int> Q;Q.push(S);dis[S]=0;vis[S]=true;
	while(!Q.empty())
	{
		int u=Q.front();Q.pop();
		for(int i=h[u];i;i=e[i].next)
			if(dis[e[i].v]>dis[u]+e[i].w)
			{
				dis[e[i].v]=dis[u]+e[i].w;
				if(!vis[e[i].v])vis[e[i].v]=true,Q.push(e[i].v);
			}
		vis[u]=false;
	}
	return dis[T]<1e9?dis[T]:-1;
}
int main()
{
	n=read();m=read();Q=read();
	scanf("%s",dS+1);scanf("%s",dT+1);
	for(int i=1;i<=n;++i)
		if(dS[i]=='E')E.push_back(i);
		else W.push_back(i);
	for(int i=1;i<=m;++i)
		if(dT[i]=='N')N.push_back(i);
		else S.push_back(i);
	while(Q--)
	{
		int a=read(),b=read(),c=read(),d=read();
		vector<int> X,Y;
		X.push_back(Findpre(E,a));X.push_back(Findpre(W,a));
		X.push_back(Findsuf(E,a));X.push_back(Findsuf(W,a));
		X.push_back(Findpre(E,c));X.push_back(Findpre(W,c));
		X.push_back(Findsuf(E,c));X.push_back(Findsuf(W,c));
		Y.push_back(Findpre(N,b));Y.push_back(Findpre(S,b));
		Y.push_back(Findsuf(N,b));Y.push_back(Findsuf(S,b));
		Y.push_back(Findpre(N,d));Y.push_back(Findpre(S,d));
		Y.push_back(Findsuf(N,d));Y.push_back(Findsuf(S,d));
		sort(X.begin(),X.end());X.resize(unique(X.begin(),X.end())-X.begin());
		sort(Y.begin(),Y.end());Y.resize(unique(Y.begin(),Y.end())-Y.begin());
		int lx=X.size(),ly=Y.size(),S,T;tot=0;
		for(int i=0;i<lx;++i)
			for(int j=0;j<ly;++j)
			{
				bh[i][j]=++tot;
				if(X[i]==a&&Y[j]==b)S=tot;
				if(X[i]==c&&Y[j]==d)T=tot;
			}
		for(int i=1;i<=tot;++i)h[i]=0;cnt=2;
		for(int i=0;i<lx;++i)
			if(dS[X[i]]=='E')for(int j=0;j<ly-1;++j)Add(bh[i][j],bh[i][j+1],Y[j+1]-Y[j]);
			else for(int j=1;j<ly;++j)Add(bh[i][j],bh[i][j-1],Y[j]-Y[j-1]);
		for(int i=0;i<ly;++i)
			if(dT[Y[i]]=='N')for(int j=1;j<lx;++j)Add(bh[j][i],bh[j-1][i],X[j]-X[j-1]);
			else for(int j=0;j<lx-1;++j)Add(bh[j][i],bh[j+1][i],X[j+1]-X[j]);
		printf("%d\n",SPFA(S,T));
	}
}