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;
#define nn 400008
#define pii pair<int,int>
#define mp make_pair
int n,m;int q;pii from[nn],dest[nn];
int ans[nn];
int dir[2][nn];
vector<int> path[2][2];
/*
0...
....
...1

0->
1|
*/
void bb(vector<int> &v)
{
	sort(v.begin(),v.end());
	v.resize(unique(v.begin(),v.end())-v.begin());
}
int lb(int x,vector<int> &v) {if(v.size()==0 or v[0]>x) return x;return v[upper_bound(v.begin(),v.end(),x)-v.begin()-1];}
int ub(int x,vector<int> &v) {if(v.size()==0 or v[v.size()-1]<x) return x;return v[lower_bound(v.begin(),v.end(),x)-v.begin()];}
int trcyx(pii a,pii b) {return abs(a.first-b.first)+abs(a.second-b.second);}

int tot;
vector<int> v[nn],c[nn];
void clear()
{
	for(int i=0;i<=tot;i++) v[i].clear(),c[i].clear();
	tot=0;
}
void link(int x,int y,int w)
{
	v[x].push_back(y);c[x].push_back(w);
}


int dist[nn];
int dj(int x,int y)
{
	for(int i=1;i<=tot;++i) dist[i]=1e9;
	dist[x]=0;priority_queue<pii> pq;pq.push(mp(-dist[x],x));
	while(pq.size()) 
	{
		int z=pq.top().second;//if(pq.top().first!=-dist[x]) {pq.pop();continue;}
		pq.pop();
		for(int i=0;i<v[z].size();i++)
		{
			int w=v[z][i];
			if(dist[w]>dist[z]+c[z][i])
			{
				dist[w]=dist[z]+c[z][i];
				pq.push(mp(-dist[w],w));
			}
		}
	}
	return dist[y];
}
void solve()
{
	for(int i=1;i<=n;i++)
		path[0][dir[0][i]].push_back(i);
	for(int i=1;i<=m;i++)
		path[1][dir[1][i]].push_back(i);
	
	for(int i=1;i<=q;i++)
	{
		clear();
		vector<int> allx,ally;int x,y;
		
		x=from[i].first,y=from[i].second;
		for(int j=0;j<2;j++) allx.push_back(lb(x,path[0][j])),allx.push_back(ub(x,path[0][j]));
		for(int j=0;j<2;j++) ally.push_back(lb(y,path[1][j])),ally.push_back(ub(y,path[1][j]));
		
		
		x=dest[i].first,y=dest[i].second;
		for(int j=0;j<2;j++) allx.push_back(lb(x,path[0][j])),allx.push_back(ub(x,path[0][j]));
		for(int j=0;j<2;j++) ally.push_back(lb(y,path[1][j])),ally.push_back(ub(y,path[1][j]));
		
		
		sort(allx.begin(),allx.end());allx.resize(unique(allx.begin(),allx.end())-allx.begin());
		sort(ally.begin(),ally.end());ally.resize(unique(ally.begin(),ally.end())-ally.begin());
		
		
		map<pii,int> id;
		for(int i=0;i<allx.size();i++)
			for(int j=0;j<ally.size();j++)
				id[mp(allx[i],ally[j])]=++tot;
		
		for(int i=0;i<allx.size();i++)
			for(int j=1;j<ally.size();j++)
			{
				pii a=mp(allx[i],ally[j-1]),b=mp(allx[i],ally[j]);
				if(!dir[0][allx[i]]) swap(a,b);
				link(id[a],id[b],trcyx(a,b));
			}
			
		for(int j=0;j<ally.size();j++)
			for(int i=1;i<allx.size();i++)
			{
				pii a=mp(allx[i-1],ally[j]),b=mp(allx[i],ally[j]);
				if(!dir[1][ally[j]]) swap(a,b);
				link(id[a],id[b],trcyx(a,b));
			}
		
		ans[i]=dj(id[from[i]],id[dest[i]]);
	}
	
}
char buf[nn];
int main()
{
	scanf("%d%d%d",&n,&m,&q);
	scanf("%s",buf+1);for(int i=1;i<=n;i++) dir[0][i]=(buf[i]=='E');
	scanf("%s",buf+1);for(int i=1;i<=m;i++) dir[1][i]=(buf[i]=='S');
	
	for(int i=1;i<=q;i++) ans[i]=1e9,scanf("%d%d%d%d",&from[i].first,&from[i].second,&dest[i].first,&dest[i].second);	
	solve();
	
	for(int i=1;i<=q;i++) if(ans[i]<1e7)printf("%d\n",ans[i]);else printf("%d\n",-1);
	return 0;
}