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 ll long long
#define all(aaa) aaa.begin(), aaa.end()

const int dir[2] = {-1, 1};
const int N = 1e5 + 5, K = 10, INF = 1e9;
int rt_s[N][2], lt_s[N][2], rt_t[N][2], lt_t[N][2], dt[K][K];


void build(const string &z, int lt[N][2], int rt[N][2]) {
    int n = z.size();

    lt[0][0] = lt[0][1] = -1;
    rt[n - 1][0] = rt[n - 1][1] = -1;

    for (int i = 0; i < n; i++) {
        lt[i][z[i]] = i;
        if (i + 1 < n) {
            lt[i + 1][0] = lt[i][0];
            lt[i + 1][1] = lt[i][1];
        }
    }

    for (int i = n - 1; i >= 0; i--) {
        rt[i][z[i]] = i;
        if (i > 0) {
            rt[i - 1][0] = rt[i][0];
            rt[i - 1][1] = rt[i][1];
        }
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    string s, t;
    int n, m, q;
    cin >> n >> m >> q >> s >> t;

    for (char &c : s) {
        c = (c == 'E');
    }

    for (char &c : t) {
        c = (c == 'S');
    }

    build(s, lt_s, rt_s);
    build(t, lt_t, rt_t);

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

        a--, b--, c--, d--;

        vector<int> hor_lines, ver_lines;

        for (int x : {a, c}) {
            for (int i = 0; i < 2; i++) {
                if (lt_s[x][i] != -1)
                    hor_lines.push_back(lt_s[x][i]);
                if (rt_s[x][i] != -1)
                    hor_lines.push_back(rt_s[x][i]);
            }
        }

        for (int x : {b, d}) {
            for (int i = 0; i < 2; i++) {
                if (lt_t[x][i] != -1)
                    ver_lines.push_back(lt_t[x][i]);
                if (rt_t[x][i] != -1)
                    ver_lines.push_back(rt_t[x][i]);
            }
        }

        sort(all(hor_lines));
        sort(all(ver_lines));
        hor_lines.erase(unique(all(hor_lines)), hor_lines.end());
        ver_lines.erase(unique(all(ver_lines)), ver_lines.end());

        priority_queue<pair<ll, pair<int, int>>,
            vector<pair<ll, pair<int, int>>>,
            greater<pair<ll, pair<int, int>>>> pq;

        int ch = hor_lines.size(),
            cv = ver_lines.size(),
            fx, fy;

        for (int x = 0; x < ch; x++) {
            for (int y = 0; y < cv; y++) {
                if (hor_lines[x] == a &&
                    ver_lines[y] == b) {
                    dt[x][y] = 0;
                    pq.push({0, {x, y}});
                }
                else {
                    if (hor_lines[x] == c &&
                        ver_lines[y] == d) {
                        fx = x;
                        fy = y;
                    }
                    dt[x][y] = INF;
                }
            }
        }

        // for (int x : hor_lines)
        //     cout << x << " ";
        // cout << "\n";
        // for (int x : ver_lines)
        //     cout << x << " ";
        // cout << "\n";

        while (!pq.empty()) {
            int cd = pq.top().first,
                x = pq.top().second.first,
                y = pq.top().second.second;

            pq.pop();

            if (dt[x][y] != cd)
                continue;

            // cout << hor_lines[x] << " " << ver_lines[y] << " " << cd << "\n";

            // cout << ver_lines[y] << "\n";

            for (int i = 0; i < 2; i++) {
                int nx = x + ((i & 1) ? dir[t[ver_lines[y]]] : 0), 
                    ny = y + (!(i & 1) ? dir[s[hor_lines[x]]] : 0),
                    w = abs(hor_lines[x] - hor_lines[nx]) 
                        + abs(ver_lines[y] - ver_lines[ny]);

                if (nx >= 0 && nx < ch && ny >= 0 && ny < cv &&
                    dt[nx][ny] > dt[x][y] + w) {
                    dt[nx][ny] = dt[x][y] + w;
                    pq.push({dt[nx][ny], {nx, ny}});
                }
            }
        }

        if (dt[fx][fy] == INF) {
            cout << "-1\n";
        }
        else {
            cout << dt[fx][fy] << "\n";
        }
    }

    return 0;
}