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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

const int INF = INT_MAX/2;
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
using P = pair<int,int>;

vector<int> dx = {1,0,-1,0,1,1,-1,-1};
vector<int> dy = {0,1,0,-1,1,-1,1,-1};

int main() {

  int H,W,K; cin >> H >> W >> K;
  P s,g; cin >> s.first >> s.second >> g.first >> g.second;
  s.first--;s.second--;g.first--;g.second--;

  vector<string> C(H);
  REP(i,H) cin >> C.at(i);

  vector<vector<int>> len(H,vector<int>(W,INF));

  REP(i,H)REP(j,W){
    if(C.at(i).at(j)=='@')len.at(i).at(j)=-1;
  }

  len.at(s.first).at(s.second) = 0;

  queue<P> qu;

  qu.emplace(s.first,s.second);

  while(!qu.empty()){
    P now = qu.front();qu.pop();
    int a = now.first, b = now.second;
    int lengh = len.at(a).at(b);
    REP(i,4){
      for(int t = 1;t<=K;t++){
        int x = a + dx.at(i)*t, y = b + dy.at(i)*t;
        if(x<0||H<=x||y<0||W<=y)break;
        
        if(len.at(x).at(y)>lengh){
          if(chmin(len.at(x).at(y),lengh+1)) qu.emplace(x,y);
        }else{
          break;
        }
        
      }
    }
  }
  int ans = len.at(g.first).at(g.second);
  if(ans==INF) ans=-1;
  
  cout << ans << endl;

  return 0;
}