Problem: Snuke, a water strider, lives in a rectangular pond that can be seen as a grid with H east-west rows and W north-south columns. Let (i,j) be the square at the i-th row from the north and j-th column from the west.

Some of the squares have a lotus leaf on it and cannot be entered. The square (i,j) has a lotus leaf on it if c_{ij} is `@`, and it does not if c_{ij} is `.`.

In one stroke, Snuke can move between 1 and K squares (inclusive) toward one of the four directions: north, east, south, and west. The move may not pass through a square with a lotus leaf. Moving to such a square or out of the pond is also forbidden.

Find the minimum number of strokes Snuke takes to travel from the square (x_1,y_1) to (x_2,y_2). If the travel from (x_1,y_1) to (x_2,y_2) is impossible, point out that fact.

Constraints

* 1 \leq H,W,K \leq 10^6
* H \times W \leq 10^6
* 1 \leq x_1,x_2 \leq H
* 1 \leq y_1,y_2 \leq W
* x_1 \neq x_2 or y_1 \neq y_2.
* c_{i,j} is `.` or `@`.
* c_{x_1,y_1} = `.`
* c_{x_2,y_2} = `.`
* All numbers in input are integers.

Input

Input is given from Standard Input in the following format:


H W K
x_1 y_1 x_2 y_2
c_{1,1}c_{1,2} .. c_{1,W}
c_{2,1}c_{2,2} .. c_{2,W}
:
c_{H,1}c_{H,2} .. c_{H,W}


Output

Print the minimum number of strokes Snuke takes to travel from the square (x_1,y_1) to (x_2,y_2), or print `-1` if the travel is impossible.

Examples

Input

3 5 2
3 2 3 4
.....
.@..@
..@..


Output

5


Input

1 6 4
1 1 1 6
......


Output

2


Input

3 3 1
2 1 2 3
.@.
.@.
.@.


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

int main() {
  ll h, w, k;
  cin >> h >> w >> k;
  ll x1, y1, x2, y2;
  cin >> x1 >> y1 >> x2 >> y2;
  x1--, x2--, y1--, y2--;
  vector<string> s(h);
  REP(i, h) { cin >> s[i]; }
  vector<vector<ll>> dist(h, vector<ll>(w, -1));
  queue<P> q;
  q.push({x1, y1});
  dist[x1][y1] = 0;
  vector<ll> d = {0, 1, 0, -1};

  while (!q.empty()) {
    P p = q.front();
    q.pop();
    ll cx = p.first, cy = p.second;

    REP(i, 4) {
      FOR(j, 1, k + 1) {
        ll nx = cx + j * d[i], ny = cy + j * d[i ^ 1];
        if (nx < 0 || nx >= h || ny < 0 || ny >= w) {
          break;
        }
        if (s[nx][ny] == '@') {
          break;
        }
        if (dist[nx][ny] != -1 && dist[nx][ny] <= dist[cx][cy]) {
          break;
        }

        if (dist[nx][ny] == -1) {
          dist[nx][ny] = dist[cx][cy] + 1;
          q.push({nx, ny});
        }
      }
    }
  }
  cout << dist[x2][y2] << endl;
}