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 <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
#include <string>
#include <vector>

#define MAX_N 100000
#define MAX_M 100000

using namespace std;

int N, M, Q;
string S, T;

int lowerX[MAX_M];
int upperX[MAX_M];
int lowerY[MAX_N];
int upperY[MAX_N];

void initCompress()
{
    int le = N;
    int lw = N;
    for ( int i = 0; i < N; i++ )
    {
        if ( S[i] == 'E' ) le = i;
        if ( S[i] == 'W' ) lw = i;

        lowerY[i] = min( ( S[i] == 'E' ? lw : le ), i );
    }

    int ue = -1;
    int uw = -1;
    for ( int i = N - 1; i >= 0; i-- )
    {
        if ( S[i] == 'E' ) ue = i;
        if ( S[i] == 'W' ) uw = i;

        upperY[i] = max( ( S[i] == 'E' ? uw : ue ), i );
    }

    int ln = N;
    int ls = N;
    for ( int i = 0; i < M; i++ )
    {
        if ( T[i] == 'N' ) ln = i;
        if ( T[i] == 'S' ) ls = i;

        lowerX[i] = min( ( T[i] == 'N' ? ls : ln ), i );
    }

    int un = -1;
    int us = -1;
    for ( int i = M - 1; i >= 0; i-- )
    {
        if ( T[i] == 'N' ) un = i;
        if ( T[i] == 'S' ) us = i;

        upperX[i] = max( ( T[i] == 'N' ? us : un ), i );
    }
}

long long dist[6][6];

typedef pair<int, int> P2;
typedef pair<int, P2> P3;

int query( int y1, int x1, int y2, int x2 )
{
    vector<int> X;
    vector<int> Y;

    memset( dist, -1, sizeof( dist ) );

    // initialize X
    X.push_back( x1 );
    X.push_back( lowerX[x1] );
    X.push_back( upperX[x1] );
    X.push_back( x2 );
    X.push_back( lowerX[x2] );
    X.push_back( upperX[x2] );

    sort( X.begin(), X.end() );
    X.erase( unique( X.begin(), X.end() ), X.end() );

    // initialize Y
    Y.push_back( y1 );
    Y.push_back( lowerY[y1] );
    Y.push_back( upperY[y1] );
    Y.push_back( y2 );
    Y.push_back( lowerY[y2] );
    Y.push_back( upperY[y2] );

    sort( Y.begin(), Y.end() );
    Y.erase( unique( Y.begin(), Y.end() ), Y.end() );

    int scx = find( X.begin(), X.end(), x1 ) - X.begin(); // index of start x
    int scy = find( Y.begin(), Y.end(), y1 ) - Y.begin(); // index of start y
    int gcx = find( X.begin(), X.end(), x2 ) - X.begin(); // index of goal x
    int gcy = find( Y.begin(), Y.end(), y2 ) - Y.begin(); // index of goal y

    priority_queue<P3> q;
    q.push( P3( 0, P2( scx, scy ) ) );

    while ( !q.empty() )
    {
        long long d = -q.top().first;
        int cxi = q.top().second.first;
        int cyi = q.top().second.second;

        q.pop();

        if ( cxi == gcx && cyi == gcy ) return d;

        int x = X[cxi];
        int y = Y[cyi];

        if ( T[x] == 'N' && 0 <= cyi - 1 )
        {
            int ny = Y[cyi - 1];
            long long nd = d + abs( ny - y );

            if ( nd < dist[cxi][cyi - 1] || dist[cxi][cyi - 1] == -1 )
            {
                dist[cxi][cyi - 1] = nd;
                q.push( P3( -nd, P2( cxi, cyi - 1 ) ) );
            }
        }
        else if ( T[x] == 'S' && cyi + 1 < Y.size() )
        {
            int ny = Y[cyi + 1];
            long long nd = d + abs( ny - y );

            if ( nd < dist[cxi][cyi + 1] || dist[cxi][cyi + 1] == -1 )
            {
                dist[cxi][cyi + 1] = nd;
                q.push( P3( -nd, P2( cxi, cyi + 1 ) ) );
            }
        }

        if ( S[y] == 'W' && 0 <= cxi - 1 )
        {
            int nx = X[cxi - 1];
            long long nd = d + abs( nx - x );

            if ( nd < dist[cxi - 1][cyi] || dist[cxi - 1][cyi] == -1 )
            {
                dist[cxi - 1][cyi] = nd;
                q.push( P3( -nd, P2( cxi - 1, cyi ) ) );
            }
        }
        else if ( S[y] == 'E' && cxi + 1 < X.size() )
        {
            int nx = X[cxi + 1];
            long long nd = d + abs( nx - x );

            if ( nd < dist[cxi + 1][cyi] || dist[cxi + 1][cyi] == -1 )
            {
                dist[cxi + 1][cyi] = nd;
                q.push( P3( -nd, P2( cxi + 1, cyi ) ) );
            }
        }
    }

    return dist[gcx][gcy];
}

int main( int argc, char **argv )
{
    cin >> N >> M >> Q;
    cin >> S;
    cin >> T;

    initCompress();

    for ( int q = 0; q < Q; q++ )
    {
        int a, b, c, d;
        cin >> a >> b >> c >> d;

        int ans = query( a - 1, b - 1, c - 1, d - 1 );
        cout << ans << endl;
    }

    return 0;
}