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 mp make_pair
#define pb push_back
#define fi first
#define se second
#define sort stable_sort
#define all(x) (x).begin(),(x).end()
#define rep(i,x) for (int (i)=(1);(i)<=(x);(i)++)
#define rep2(i,x) for (int (i)=(x);(i)>=(1);(i)--)
#define foreach(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
using namespace std;
int n,m,q,i,j,nxtx[100005],prex[100005],nxty[100005],prey[100005],x,y,xx,yy,lst1,lst2;
ll dis[55];
string s,t;
char ss[100005];
vector<pair<int,int> > all;
vector<pair<int,int> > bi[55];
void add(int x,int y)
{
	all.push_back(make_pair(x,y));
	if (prex[x]) all.push_back(make_pair(prex[x],y));
	if (nxtx[x]) all.push_back(make_pair(nxtx[x],y));
	if (prey[y]) all.push_back(make_pair(x,prey[y]));
	if (nxty[y]) all.push_back(make_pair(x,nxty[y]));
	if (prex[x]&&prey[y]) all.push_back(make_pair(prex[x],prey[y]));
	if (prex[x]&&nxty[y]) all.push_back(make_pair(prex[x],nxty[y]));
	if (nxtx[x]&&prey[y]) all.push_back(make_pair(nxtx[x],prey[y]));
	if (nxtx[x]&&nxty[y]) all.push_back(make_pair(nxtx[x],nxty[y]));
}
ll dijkstra(int x,int y)
{
	memset(dis,18,sizeof(dis));
	dis[x]=0;
	priority_queue<pair<ll,ll> > pq;
	pq.push(make_pair(0,x));
	while (!pq.empty())
	{
		x=pq.top().second;pq.pop();
		foreach(it,bi[x])
		{
			if (dis[it->fi]>dis[x]+it->se)
			{
				dis[it->fi]=dis[x]+it->se;
				pq.push(make_pair(-dis[it->fi],it->fi));
			}
		}
	}
	if (dis[y]==dis[54]) return -1;
	return dis[y];
}
void solve(int x,int y,int xx,int yy)
{
	sort(all.begin(),all.end());
	all.resize(unique(all.begin(),all.end())-all.begin());
//	foreach(it,all) cerr<<it->fi<<' '<<it->se<<endl;
	int i,j,ss,tt;
	for (i=0;i<all.size();i++) if (all[i].first==x&&all[i].second==y) ss=i;
	for (i=0;i<all.size();i++) if (all[i].first==xx&&all[i].second==yy) tt=i;
	for (i=0;i<all.size();i++) bi[i].clear();
	for (i=0;i<all.size();i++) for (j=0;j<all.size();j++) if (i!=j)
	{
		if (all[i].first==all[j].first&&((s[all[i].first]=='E')==(all[i].second<all[j].second)))
		{
			bi[i].push_back(make_pair(j,abs(all[i].second-all[j].second)));
		}
		if (all[i].second==all[j].second&&((t[all[i].second]=='S')==(all[i].first<all[j].first)))
		{
			bi[i].push_back(make_pair(j,abs(all[i].first-all[j].first)));
		}
	}
	printf("%lld\n",dijkstra(ss,tt));
}
int main()
{
	scanf("%d%d%d",&n,&m,&q);
	scanf(" %s",ss);s=ss;s=" "+s;
	scanf(" %s",ss);t=ss;t=" "+t;
	lst1=lst2=0;
	rep(i,n){if (s[i]=='W'){prex[i]=lst2;lst1=i;}else{prex[i]=lst1;lst2=i;}}
	lst1=lst2=0;
	rep2(i,n){if (s[i]=='W'){nxtx[i]=lst2;lst1=i;}else{nxtx[i]=lst1;lst2=i;}}
	lst1=lst2=0;
	rep(i,m){if (t[i]=='N'){prey[i]=lst2;lst1=i;}else{prey[i]=lst1;lst2=i;}}
	lst1=lst2=0;
	rep2(i,m){if (t[i]=='N'){nxty[i]=lst2;lst1=i;}else{nxty[i]=lst1;lst2=i;}}
	while (q--)
	{
		scanf("%d%d%d%d",&x,&y,&xx,&yy);
		all.clear();
		add(x,y);add(x,yy);add(xx,y);add(xx,yy);
		solve(x,y,xx,yy);
	}
	return 0;
}