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:
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")

#include <bits/stdc++.h>

#define pb push_back
#define fi first
#define se second
#define all(v) v.begin(), v.end()
#define sz(v) int(v.size())

using namespace std;
using ll = int64_t;

int N, M, Q;
string S, T;
vector<int> xPositions[2];
vector<int> yPositions[2];
vector<int> interestingX;
vector<int> interestingY;
queue<pair<int,int>> q;

int distances[10][10];

void build() {
  for (int i = 0; i < N; i++)
    xPositions[S[i] == 'E' ? 1 : 0].pb(i);
  for (int i = 0; i < M; i++)
    yPositions[T[i] == 'S' ? 1 : 0].pb(i);
}

int Next(const vector<int>& v, int x) {
  auto it = lower_bound(all(v), x);
  if (it == v.end())
    return -1;
  return *it;
}

int Prev(const vector<int>& v, int x) {
  auto it = upper_bound(all(v), x);
  if (it == v.begin())
    return -1;
  --it;
  return *it;
}

void filter(vector<int>& v) {
  sort(all(v));
  v.erase(unique(all(v)), v.end());
  while (!v.empty() && v.front() == -1)
    v.erase(v.begin());
}

int ans(int x0, int y0, int x1, int y1) {
  interestingX.clear();
  interestingY.clear();
  
  interestingX.pb(x0);
  interestingX.pb(x1);
  interestingY.pb(y0);
  interestingY.pb(y1);
  for (int d = 0; d <= 1; d++) {
    interestingX.pb(Next(xPositions[d], x0));
    interestingX.pb(Prev(xPositions[d], x0));
    interestingX.pb(Next(xPositions[d], x1));
    interestingX.pb(Prev(xPositions[d], x1));

    interestingY.pb(Next(yPositions[d], y0));
    interestingY.pb(Prev(yPositions[d], y0));
    interestingY.pb(Next(yPositions[d], y1));
    interestingY.pb(Prev(yPositions[d], y1));
  }

  filter(interestingX);
  filter(interestingY);

  auto getX = [&](int x) {
    return lower_bound(all(interestingX), x) - interestingX.begin();
  };

  auto getY = [&](int y) {
    return lower_bound(all(interestingY), y) - interestingY.begin();
  };

  memset(distances, -1, sizeof(distances));
  x0 = getX(x0);
  y0 = getY(y0);
  x1 = getX(x1);
  y1 = getY(y1);
  distances[x0][y0] = 0;
  q.push({x0, y0});

  while (!q.empty()) {
    auto p = q.front();
    q.pop();
    int d = distances[p.fi][p.se];
    // cerr << "debug: " << p.fi << " " << p.se << " " << d << endl;

    // go x
    int lx = 0, rx = sz(interestingX) - 1;
    if (T[interestingY[p.se]] == 'S') lx = p.fi + 1;
    if (T[interestingY[p.se]] == 'N') rx = p.fi - 1;
    for (int nx = lx; nx <= rx; nx++) {
      int nd = d + abs(interestingX[nx] - interestingX[p.fi]);
      if (distances[nx][p.se] == -1 || distances[nx][p.se] > nd) {
        distances[nx][p.se] = nd;
        q.push({nx, p.se});
      }
    }

    // go y
    int ly = 0, ry = sz(interestingY) - 1;
    if (S[interestingX[p.fi]] == 'E') ly = p.se + 1;
    if (S[interestingX[p.fi]] == 'W') ry = p.se - 1;
    for (int ny = ly; ny <= ry; ny++) {
      int nd = d + abs(interestingY[ny] - interestingY[p.se]);
      if (distances[p.fi][ny] == -1 || distances[p.fi][ny] > nd) {
        distances[p.fi][ny] = nd;
        q.push({p.fi, ny});
      }
    }
  }

  return distances[x1][y1];
}

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

  cin >> N >> M >> Q;
  cin >> S >> T;

  build();

  for (int i = 0; i < Q; i++) {
    int x0, y0, x1, y1;
    cin >> x0 >> y0 >> x1 >> y1;
    --x0, --y0, --x1, --y1;
    cout << ans(x0, y0, x1, y1) << "\n";
  }

  return 0;
}