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;
int n,m,q,all,dist[1111],a,b,c,d,it,R,C,aa,bb,cc,dd,ans;
char s[111111],t[111111];
vector<int> h0,h1,v0,v1;
bool h[111111],v[111111];
vector<pair<int,int> > g[1111];
vector<int> ho,ve;
void dijkstra(int S)
{
	priority_queue<pair<int,int> > pq;
	for (int i=1;i<=all;i++) dist[i]=1e9;
	dist[S]=0;
	pq.push(make_pair(0,S));
	while(!pq.empty()) 
	{
		int x=pq.top().second,y=-pq.top().first;pq.pop();
		if (dist[x]!=y) continue;
		for (int i=0;i<g[x].size();i++)
		{
			int to=g[x][i].first,val=g[x][i].second;
			if (dist[to]>dist[x]+val)
			{
				dist[to]=dist[x]+val;
				pq.push(make_pair(-dist[to],to)); 
			}
		}
	}
}
int main()
{
	scanf("%d%d%d",&n,&m,&q);
	scanf("%s%s",s,t);
	for (int i=0;i<n;i++)
	{
		h[i]=(s[i]=='W');
	}
	for (int i=0;i<m;i++)
	{
		v[i]=(t[i]=='N');
	}
	for (int i=0;i<n;i++)
	{
		if (h[i])
		{
			h1.push_back(i); 
		}
		else 
		{
			h0.push_back(i); 
		}
	}
	for (int i=0;i<m;i++)
	{
		if (v[i])
		{
			v1.push_back(i); 
		}
		else
		{
			v0.push_back(i); 
		}
	}
	while(q--)
	{
		scanf("%d%d%d%d",&a,&b,&c,&d);
		a--;b--;c--;d--;
		ho.clear();ve.clear();
		ho.push_back(a);
		ho.push_back(c);
		ve.push_back(b);
		ve.push_back(d);    
		if (h[a])
		{
			it=lower_bound(h0.begin(),h0.end(),a)-h0.begin();
			if (it<(int)h0.size()) ho.push_back(h0[it]); 
			it--;
			if (it>=0) ho.push_back(h0[it]); 
		} 
		else
		{
			it=lower_bound(h1.begin(),h1.end(),a)-h1.begin();
			if (it<(int)h1.size()) ho.push_back(h1[it]); 
			it--;
			if (it>=0) ho.push_back(h1[it]); 
		}
		if (h[c])
		{
			it=lower_bound(h0.begin(),h0.end(),c)-h0.begin();
			if (it<(int)h0.size()) ho.push_back(h0[it]); 
			it--;
			if (it>=0) ho.push_back(h0[it]); 
		} 
		else
		{
			it=lower_bound(h1.begin(),h1.end(),c)-h1.begin();
			if (it<(int)h1.size()) ho.push_back(h1[it]); 
			it--;
			if (it>=0) ho.push_back(h1[it]); 
		}
		if (v[b])
		{
			it=lower_bound(v0.begin(),v0.end(),b)-v0.begin();
			if (it<(int)v0.size()) ve.push_back(v0[it]);
			it--;
			if (it>=0) ve.push_back(v0[it]);  
		}
		else
		{
			it=lower_bound(v1.begin(),v1.end(),b)-v1.begin();
			if (it<(int)v1.size()) ve.push_back(v1[it]);
			it--;
			if (it>=0) ve.push_back(v1[it]); 
		}
		if (v[d])
		{
			it=lower_bound(v0.begin(),v0.end(),d)-v0.begin();
			if (it<(int)v0.size()) ve.push_back(v0[it]);
			it--;
			if (it>=0) ve.push_back(v0[it]);  
		}
		else
		{
			it=lower_bound(v1.begin(),v1.end(),d)-v1.begin();
			if (it<(int)v1.size()) ve.push_back(v1[it]);
			it--;
			if (it>=0) ve.push_back(v1[it]); 
		}
		sort(ho.begin(),ho.end());
		sort(ve.begin(),ve.end());
		ho.resize(unique(ho.begin(),ho.end())-ho.begin());
		ve.resize(unique(ve.begin(),ve.end())-ve.begin());
		R=ho.size();C=ve.size();
		//for (int i=0;i<R;i++) cout<<ho[i]<<" ";
		//cout<<endl;
		//for (int i=0;i<C;i++) cout<<ve[i]<<" ";
		//cout<<endl;
		all=R*C;
		for (int i=1;i<=all;i++) g[i].clear();
		for (int i=0;i<R;i++)
		{
			for (int j=0;j<C-1;j++)
			{
				if (h[ho[i]])
				{
					g[i*C+j+2].push_back(make_pair(i*C+j+1,ve[j+1]-ve[j])); 
				}
				else
				{
					g[i*C+j+1].push_back(make_pair(i*C+j+2,ve[j+1]-ve[j])); 
				}
			}
		}
		for (int j=0;j<C;j++)
		{
			for (int i=0;i<R-1;i++)
			{
				if (v[ve[j]]) 
				{
					g[(i+1)*C+j+1].push_back(make_pair(i*C+j+1,ho[i+1]-ho[i])); 
				}
				else
				{
					g[i*C+j+1].push_back(make_pair((i+1)*C+j+1,ho[i+1]-ho[i])); 
				}
			}
		}
		for (int i=0;i<R;i++)
		{
			if (ho[i]==a) aa=i;
			if (ho[i]==c) cc=i;
		}
		for (int i=0;i<C;i++)
		{
			if (ve[i]==b) bb=i;
			if (ve[i]==d) dd=i;
		}
		dijkstra(aa*C+bb+1);
		ans=dist[cc*C+dd+1];
		if (ans>1e8) printf("-1\n");
		else printf("%d\n",ans);
	}
	return 0;
}