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"
#define pb push_back
#define pii pair<int,int>
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define F first
#define S second
#define LINE cout<<"--------------------------\n";
#define IOS ios::sync_with_stdio(false),cin.tie(0);
using namespace std;
int dx[]={0 , 0 , -1 , 1} , dy[]={1 , -1 , 0 , 0};
signed main(){
	IOS;
	int n , m , k ; cin >> n >> m >> k;
	int x , y , x2 , y2; cin >> x >> y >> x2 >> y2 , --x , --y , --x2 , --y2;
	vector<string> c(n);
	vector<vector<int>> dis(n,vector<int>(m,1e9));
	for(int i=0 ; i<n ; ++i) cin >> c[i];
	queue<pii> pq;
	if(c[x][y] != '@')pq.push({x , y}) , dis[x][y] = 0;
	while(pq.size()){
		int x = pq.front().F , y = pq.front().S , d = dis[x][y]+1; pq.pop();
		for(int i=0 ; i<4 ; ++i){
			int xx = x , yy = y;
			for(int _ = 1;  _<=k ; ++_){
				xx += dx[i] , yy += dy[i];
				if(xx<0 || xx>=n || yy<0 || yy>=m || c[xx][yy]=='@' || dis[xx][yy] < d) break;
				if(dis[xx][yy] > d) dis[xx][yy] = d , pq.push({xx , yy});
			}
		}
	}
	cout << (dis[x2][y2]==1e9?-1:dis[x2][y2]) << '\n';
}