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;
struct state{int x,y,d,dir;}st[5][11001];
char s[110001],t[110001];
int n,m,q,sl[110001][2],sr[110001][2],tl[110001][2],tr[110001][2],ex,ey,sx,sy,ans,cnt[5];
void go(int id,int x,int y,int d,bool ns)
{
	if(ns)
		if(t[y]=='S')
		{
			if(sr[x][0])st[id][cnt[id]++]={sr[x][0],y,d+sr[x][0]-x,ns};
			if(sr[x][1])st[id][cnt[id]++]={sr[x][1],y,d+sr[x][1]-x,ns};
			if(ex>x)
			{
				if(ey==y)ans=min(ans,d+ex-x);
				if(int z=sr[ex-1][ey>y])st[id][cnt[id]++]={z,y,d+z-x,ns};
			}
		}
		else
		{
			if(sl[x][0])st[id][cnt[id]++]={sl[x][0],y,d+x-sl[x][0],ns};
			if(sl[x][1])st[id][cnt[id]++]={sl[x][1],y,d+x-sl[x][1],ns};
			if(ex<x)
			{
				if(ey==y)ans=min(ans,d+x-ex);
				if(int z=sl[ex+1][ey>y])st[id][cnt[id]++]={z,y,d+x-z,ns};
			}
		}
	else if(s[x]=='E')
	{
		if(tr[y][0])st[id][cnt[id]++]={x,tr[y][0],d+tr[y][0]-y,ns};
		if(tr[y][1])st[id][cnt[id]++]={x,tr[y][1],d+tr[y][1]-y,ns};
		if(ey>y)
		{
			if(ex==x)ans=min(ans,d+ey-y);
			if(int z=tr[ey-1][ex>x])st[id][cnt[id]++]={x,z,d+z-y,ns};
		}
	}
	else
	{
		if(tl[y][0])st[id][cnt[id]++]={x,tl[y][0],d+y-tl[y][0],ns};
		if(tl[y][1])st[id][cnt[id]++]={x,tl[y][1],d+y-tl[y][1],ns};
		if(ey<y)
		{
			if(ex==x)ans=min(ans,d+y-ey);
			if(int z=tl[ey+1][ex>x])st[id][cnt[id]++]={x,z,d+y-z,ns};
		}
	}
}
int main()
{
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n>>m>>q>>(s+1)>>(t+1);
	for(int i=2;i<=n;i++){sl[i][0]=sl[i-1][0];sl[i][1]=sl[i-1][1];sl[i][s[i-1]=='E']=i-1;}
	for(int i=2;i<=m;i++){tl[i][0]=tl[i-1][0];tl[i][1]=tl[i-1][1];tl[i][t[i-1]=='S']=i-1;}
	for(int i=n-1;i;i--){sr[i][0]=sr[i+1][0];sr[i][1]=sr[i+1][1];sr[i][s[i+1]=='E']=i+1;}
	for(int i=m-1;i;i--){tr[i][0]=tr[i+1][0];tr[i][1]=tr[i+1][1];tr[i][t[i+1]=='S']=i+1;}
	while(q--)
	{
		cin>>sx>>sy>>ex>>ey;
		ans=1<<30;
		memset(cnt,0,sizeof cnt);
		go(0,sx,sy,0,0);
		go(0,sx,sy,0,1);
		for(int i=0;i<4;i++)
			for(int j=0;j<cnt[i];j++)go(i+1,st[i][j].x,st[i][j].y,st[i][j].d,st[i][j].dir^1);
		cout<<(ans==(1<<30)?-1:ans)<<'\n';
	}
}