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 rep(i, N) for (int i = 0; i < (N); i++)
#define all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;
using ll_i = tuple<ll, int>;
struct edge { int v, w; };

vector<ll> dijkstra(int N, vector<vector<edge> >& G, int s) {
	vector<ll> d(N, LLONG_MAX); d[s] = 0;
	priority_queue<ll_i, vector<ll_i>, greater<ll_i> > pq;
	pq.push(ll_i(0, s));
	while (pq.size()) {
		ll_i p = pq.top(); pq.pop();
		int u = get<1>(p);
		if (get<0>(p) > d[u]) continue;
		for (edge e: G[u])
			if (d[e.v] > d[u] + e.w) {
				d[e.v] = d[u] + e.w;
				pq.push(ll_i(d[e.v], e.v));
			}
	}
	return d;
}

const int e5 = 1e5;

int a[e5], b[e5];

void f(int l, int r, vector<int>& from, vector<int>& to, bool mi, bool ma) {
    if (mi) {
        auto it = lower_bound(all(from), l);
        if (it != from.end()) {
            int i = *it;
            if (l <= i && i <= r) to.pb(i);
        }
    }
    if (ma) {
        auto it = upper_bound(all(from), r);
        if (it != from.begin()) {
            int i = *prev(it);
            if (l <= i && i <= r) to.pb(i);
        }
    }
}

int main() {
    int N, M, Q;
    cin >> N >> M >> Q;
    string s, t;
    cin >> s >> t;
    vector<int> a(N), b(M);
    rep(i, N) a[i] = (s[i] == 'E');
    rep(j, M) b[j] = (t[j] == 'S');
    vector<vector<int>> A(2), B(2);
    rep(i, N) A[a[i]].pb(i);
    rep(j, M) B[b[j]].pb(j);
    while (Q--) {
        int i_s, j_s, i_t, j_t;
        scanf("%d%d%d%d", &i_s, &j_s, &i_t, &j_t);
        i_s--; j_s--; i_t--; j_t--;
        vector<int> I = {-1, N, i_s, i_t};
        vector<int> J = {-1, M, j_s, j_t};
        sort(all(I)); sort(all(J));
        I.erase(unique(all(I)), I.end());
        J.erase(unique(all(J)), J.end());
        vector<int> R, C;
        for (int i: I) rep(z, 2) f(i, i, A[z], R, 0, 1);
        for (int j: J) rep(z, 2) f(j, j, B[z], C, 0, 1);
        rep(k, I.size() - 1) rep(z, 2) f(I[k] + 1, I[k + 1] - 1, A[z], R, k, !k);
        rep(k, J.size() - 1) rep(z, 2) f(J[k] + 1, J[k + 1] - 1, B[z], C, k, !k);
        sort(all(R)); sort(all(C));
        R.erase(unique(all(R)), R.end());
        C.erase(unique(all(C)), C.end());
        /*
        for (int y: R) cout << y << ' ';
        cout << endl;
        for (int x: C) cout << x << ' ';
        cout << endl;
        */
        int H = R.size(), W = C.size();
        vector<vector<edge>> G(H * W);
        rep(i, H) rep(j, W - 1) {
            int y = R[i];
            if (a[y]) G[i * W + j].pb(edge{i * W + j + 1, C[j + 1] - C[j]});
            if (!a[y]) G[i * W + j + 1].pb(edge{i * W + j, C[j + 1] - C[j]});
        }
        rep(j, W) rep(i, H - 1) {
            int x = C[j];
            if (b[x]) G[i * W + j].pb(edge{(i + 1) * W + j, R[i + 1] - R[i]});
            if (!b[x]) G[(i + 1) * W + j].pb(edge{i * W + j, R[i + 1] - R[i]});
        }
        int y_s = lower_bound(all(R), i_s) - R.begin();
        int x_s = lower_bound(all(C), j_s) - C.begin();
        int y_t = lower_bound(all(R), i_t) - R.begin();
        int x_t = lower_bound(all(C), j_t) - C.begin();
        // cout << y_s << ' ' << x_s << ' ' << y_t << ' ' << x_t << endl;
        auto d = dijkstra(H * W, G, y_s * W + x_s);
        ll ans = d[y_t * W + x_t];
        if (ans == LLONG_MAX) ans = -1;
        printf("%lld\n", ans);
    }
}