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;
#define rep(i,n) for(int i=0; i<(n); i++)
using P = pair<int,int>;
constexpr int INF = 1e9+7;
int h, w, k, x1, yy1, x2, y2, dx[] = {0,0,1,-1}, dy[] = {1,-1,0,0};
vector<vector<int>> mp;
queue<P> qu;
char c[1000005];
int main(){
	scanf("%d%d%d", &h, &w, &k);
	scanf("%d%d%d%d", &x1, &yy1, &x2, &y2);
	mp.resize(h);
	x1--; yy1--; x2--; y2--;
	rep(i,h){
		mp[i].resize(w);
		scanf("\n%s", c);
		rep(j,w) mp[i][j] = c[j] == '@' ? -1 : INF;
	}
	qu.emplace(x1,yy1);
	mp[x1][yy1] = 0;
	while(!qu.empty()){
		auto [x, y] = qu.front(); qu.pop();
		rep(i,4) for(int nx = x + dx[i], ny = y + dy[i], u = 0; u < k; u++, nx += dx[i], ny += dy[i]){
			if(nx < 0 || h <= nx || ny < 0 || w <= ny || mp[nx][ny] <= mp[x][y]) break;
			if(mp[nx][ny] == INF) qu.emplace(nx, ny);
			mp[nx][ny] = mp[x][y] + 1;
		}
	}
	printf("%d\n", mp[x2][y2] == INF ? -1 : mp[x2][y2]);
}