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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const long double EPS=1e-9;
const long double PI=acos(-1.0);
int ABS(int a){return max(a,-a);}
long long ABS(long long a){return max(a,-a);}
long double ABS(long double a){return max(a,-a);}
char S[110000];
char T[110000];
int zx[10];
int zy[10];
vector<int>W;
vector<int>E;
vector<int>N;
vector<int>D;
int ijk[10][10];
int v[10][10];
int calc(int X1,int Y1,int X2,int Y2,int n,int m){
	for(int i=0;i<n;i++)for(int j=0;j<m;j++){
		v[i][j]=0;
		ijk[i][j]=mod;
	}
	//for(int i=0;i<n;i++)printf("%d ",zx[i]);printf("\n");
	//for(int i=0;i<m;i++)printf("%d ",zy[i]);printf("\n");
		
	int XX1=lower_bound(zx,zx+n,X1)-zx;
	int YY1=lower_bound(zy,zy+m,Y1)-zy;
	int XX2=lower_bound(zx,zx+n,X2)-zx;
	int YY2=lower_bound(zy,zy+m,Y2)-zy;
	ijk[XX1][YY1]=0;
	priority_queue<pair<int,pair<int,int> > >Q;
	Q.push(make_pair(0,make_pair(XX1,YY1)));
	while(Q.size()){
		int cost=-Q.top().first;
		int row=Q.top().second.first;
		int col=Q.top().second.second;
		Q.pop();
		if(v[row][col])continue;
		v[row][col]=1;
		if(S[zx[row]]=='E'){
			if(col<m-1){
				int toc=cost+zy[col+1]-zy[col];
				if(v[row][col+1]==0&&ijk[row][col+1]>toc){
					ijk[row][col+1]=toc;
					Q.push(make_pair(-toc,make_pair(row,col+1)));
				}
			}
		}else{
			if(col){
				int toc=cost+zy[col]-zy[col-1];
				if(v[row][col-1]==0&&ijk[row][col-1]>toc){
					ijk[row][col-1]=toc;
					Q.push(make_pair(-toc,make_pair(row,col-1)));
				}
			}
		}
		if(T[zy[col]]=='S'){
			if(row<n-1){
				int toc=cost+zx[row+1]-zx[row];
				if(v[row+1][col]==0&&ijk[row+1][col]>toc){
					ijk[row+1][col]=toc;
					Q.push(make_pair(-toc,make_pair(row+1,col)));
				}
			}
		}else{
			if(row){
				int toc=cost+zx[row]-zx[row-1];
				if(v[row-1][col]==0&&ijk[row-1][col]>toc){
					ijk[row-1][col]=toc;
					Q.push(make_pair(-toc,make_pair(row-1,col)));
				}
			}
		}
	}
	if(ijk[XX2][YY2]==mod)return -1;
	else return ijk[XX2][YY2];
}
int main(){
	int a,b,c;scanf("%d%d%d",&a,&b,&c);
	scanf("%s%s",S,T);
	for(int i=0;i<a;i++){
		if(S[i]=='W')W.push_back(i);
		else E.push_back(i);
	}
	for(int i=0;i<b;i++){
		if(T[i]=='N')N.push_back(i);
		else D.push_back(i);
	}
	while(c--){
		int X1,Y1,X2,Y2;
		scanf("%d%d%d%d",&X1,&Y1,&X2,&Y2);
		X1--;Y1--;X2--;Y2--;
		int xs=0;
		int ys=0;
		zx[xs++]=X1;
		zx[xs++]=X2;
		
		int at=lower_bound(W.begin(),W.end(),X1)-W.begin();
		if(at<W.size()&&W[at]==X1){
			at=lower_bound(E.begin(),E.end(),X1)-E.begin();
			if(at)zx[xs++]=E[at-1];
			if(at<E.size())zx[xs++]=E[at];
		}else{
			if(at)zx[xs++]=W[at-1];
			if(at<W.size())zx[xs++]=W[at];
		}
		at=lower_bound(W.begin(),W.end(),X2)-W.begin();
		if(at<W.size()&&W[at]==X2){
			at=lower_bound(E.begin(),E.end(),X2)-E.begin();
			if(at)zx[xs++]=E[at-1];
			if(at<E.size())zx[xs++]=E[at];
		}else{
			if(at)zx[xs++]=W[at-1];
			if(at<W.size())zx[xs++]=W[at];
		}


		zy[ys++]=Y1;
		zy[ys++]=Y2;
		at=lower_bound(N.begin(),N.end(),Y1)-N.begin();
		if(at<N.size()&&N[at]==Y1){
			at=lower_bound(D.begin(),D.end(),Y1)-D.begin();
			if(at)zy[ys++]=D[at-1];
			if(at<D.size())zy[ys++]=D[at];
		}else{
			if(at)zy[ys++]=N[at-1];
			if(at<N.size())zy[ys++]=N[at];
		}
		at=lower_bound(N.begin(),N.end(),Y2)-N.begin();
		if(at<N.size()&&N[at]==Y2){
			at=lower_bound(D.begin(),D.end(),Y2)-D.begin();
			if(at)zy[ys++]=D[at-1];
			if(at<D.size())zy[ys++]=D[at];
		}else{
			if(at)zy[ys++]=N[at-1];
			if(at<N.size())zy[ys++]=N[at];
		}
		std::sort(zx,zx+xs);
		std::sort(zy,zy+ys);
		printf("%d\n",calc(X1,Y1,X2,Y2,xs,ys));
	}
}