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 <stdio.h>
#include <algorithm>
#include <queue>
using namespace std;

struct phase{
	int x,y,d; pair<int, int> c;
  bool operator <(const phase &t) const{
    if (c.first == t.c.first) return c.second < t.c.second;
return c.first > t.c.first;
  }
};

int n,m,k;
int dx[4] = {0,1,0,-1};
int dy[4] = {1,0,-1,0};
int sx,sy, ex,ey;

char map[10001001];
pair<int, int> e[1001001][2];

priority_queue<phase> q;
void enque(int x, int y, int d, pair<int, int> c)
{
  if (e[x*m+y][d] > c){
e[x*m+y][d] = c;
    q.push({x,y,d,c});
  }
}

int main()
{
  scanf ("%d %d %d",&n,&m,&k);
  scanf ("%d %d %d %d",&sx,&sy,&ex,&ey); sx--; sy--; ex--; ey--;
  for (int i=0;i<n;i++) scanf ("%s",map+i*m);
  for (int i=0;i<n*m;i++) e[i][0] = e[i][1] = {100000000, 0};
enque(sx, sy, 0, {0,0});
enque(sx, sy, 1, {0,0});
  while (!q.empty()){
int x = q.top().x;
int y = q.top().y;
int d = q.top().d;
    auto c = q.top().c;
    q.pop();
    
    if (e[x*m+y][d] < c) continue;
    for (int f=0;f<4;f++){
int px= x+dx[f];
      int py = y + dy[f];
      if (px < 0 || px >= n || py < 0 || py >= m || map[px*m+py] == '@') continue;
      if (c.second < 0 && f % 2 == d) enque(px,py,f%2,{c.first,c.second+1});
      else enque(px,py,f%2,{c.first+1,1-k});
    }
  }
  
  int ans = min(e[ex*m+ey][0].first,e[ex*m+ey][1].first);
  if (ans == 100000000) ans = -1;
  printf ("%d\n",ans);
  
 return 0; 
}