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;

typedef long long ll;
typedef long double ld;
int n, m, q;
string s, t;
const int mod = (int)1e9 + 7;
const int maxN = (int)1e5 + 100;
int tp[2][maxN];
int nxt[2][2][2][maxN];
int num[2];
const int INF = (int)1e9;
const int BUBEN = 4;
int get_dir(int a) {
    if (a > 0) return 1;
    else if (a == 0) return 0;
    else return -1;
}
// nxt[a][b][c][d]
// a - row / cols
// b - find up or down
// c - upper or downer
// d - who
map < pair < int, int >, int > mp;
bool can_simple(int fromx, int fromy, int tox, int toy) {
    bool ok1 = true;
    bool ok2 = true;
    if (fromx != tox) {
        if (get_dir(tox - fromx) != tp[1][fromy]) {
            ok1 = false;
        }
        if (get_dir(tox - fromx) != tp[1][toy]) ok2 = false;
    }
    if (fromy != toy) {
        if (get_dir(toy - fromy) != tp[0][tox]) ok1 = false;
        if (get_dir(toy - fromy) != tp[0][fromx]) ok2 = false;
    }
    if (ok1 || ok2) return true;
    return false;
}
int solve(int fromx, int fromy, int tox, int toy, int step) {
   // cout << fromx << " " << fromy << " " << tox << " " << toy << " " << step << endl;
    /*if (fromx == 3 && fromy == 1 && step == 5) {
        exit(0);
    }*/
    if (can_simple(fromx, fromy, tox, toy)) return abs(fromx - tox) + abs(fromy - toy);
    if (step == BUBEN) return INF;
    if (mp.count(make_pair(fromx, fromy)) && mp[make_pair(fromx, fromy)] <= step) return INF;
    mp[make_pair(fromx, fromy)] = step;
    int ans = INF;
    bool ok1 = true;
    bool ok2 = true;
    if (fromx != tox) {
        if (get_dir(tox - fromx) != tp[1][fromy]) {
            ok1 = false;
        }
        if (get_dir(tox - fromx) != tp[1][toy]) ok2 = false;
    }
    if (fromy != toy) {
        if (get_dir(toy - fromy) != tp[0][tox]) ok1 = false;
        if (get_dir(toy - fromy) != tp[0][fromx]) ok2 = false;
    }
    if (ok1 || ok2) return abs(fromx - tox) + abs(fromy - toy);
    for (int i = 0; i < 2; i++) {
        for (int k = -1; k <= 1; k += 2) {
            int val = (k + 1) / 2;
            if (nxt[0][val][i][fromx] == -1) continue;
            int where = nxt[0][val][i][fromx];
            if (where == fromx) continue;
            if (where != fromx) {
                if (get_dir(where - fromx) != tp[1][fromy]) continue;
            }
            ans = min(ans, abs(where - fromx) + solve(where, fromy, tox, toy, step + 1));
        }
    }
    for (int i = 0; i < 2; i++) {
        for (int k = -1; k <= 1; k += 2) {
            int val = (k + 1) / 2;
            if (nxt[1][val][i][fromy] == -1) continue;
            int where = nxt[1][val][i][fromy];
            if (where == fromy) continue;
            if (where != fromy) {
                if (get_dir(where - fromy) != tp[0][fromx]) continue;
            }
            ans = min(ans, abs(where - fromy) + solve(fromx, where, tox, toy, step + 1));
        }
    }
    for (int i = 0; i < 2; i++) {
        for (int k = -1; k <= 1; k += 2) {
            int val = (k + 1) / 2;
            if (nxt[1][val][i][toy] == -1) continue;
            int where = nxt[1][val][i][toy];
            if (where == fromy) continue;
            if (where != fromy) {
                if (get_dir(where - fromy) != tp[0][fromx]) continue;
            }
            ans = min(ans, abs(where - fromy) + solve(fromx, where, tox, toy, step + 1));
        }
    }
    for (int i = 0; i < 2; i++) {
        for (int k = -1; k <= 1; k += 2) {
            int val = (k + 1) / 2;
            if (nxt[0][val][i][tox] == -1) continue;
            int where = nxt[0][val][i][tox];
            if (where == fromx) continue;
            if (where != fromx) {
                if (get_dir(where - fromx) != tp[1][fromy]) continue;
            }
            ans = min(ans, abs(where - fromx) + solve(where, fromy, tox, toy, step + 1));
        }
    }
    return ans;
}
signed main() {
    //freopen("input.txt", "r", stdin);
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n >> m >> q;
    num[0] = n;
    num[1] = m;
    cin >> s >> t;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == 'E') tp[0][i + 1] = 1;
        else tp[0][i + 1] = -1;
    }
    for (int j = 0; j < t.size(); j++) {
        if (t[j] == 'N') tp[1][j + 1] = -1;
        else tp[1][j + 1] = 1;
    }
    for (int i = 0; i < 2; i++) {
        for (int k = -1; k <= 1; k += 2) {
            int val = (k + 1) / 2;
            nxt[i][val][0][0] = -1;
            for (int j = 1; j <= num[i]; j++) {
                nxt[i][val][0][j] = nxt[i][val][0][j - 1];
                if (tp[i][j] == k) nxt[i][val][0][j] = j;
            }
            nxt[i][val][1][num[i] + 1] = -1;
            for (int j = num[i]; j >= 1; j--) {
                nxt[i][val][1][j] = nxt[i][val][1][j + 1];
                if (tp[i][j] == k) nxt[i][val][1][j] = j;
            }
        }
    }
    for (int i = 0; i < q; i++) {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        mp.clear();
        int x = solve(a, b, c, d, 0);
        if (x >= INF) cout << -1 << '\n';
        else cout << x << '\n';
    }
    return 0;
}