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;

int main(){
    int h, w, k;
    cin >> h >> w >> k;
    int y1, x1, y2, x2;
    cin >> y1 >> x1 >> y2 >> x2;
    y1--; x1--; y2--; x2--;

    vector<string> C(h);
    for (auto &c: C)
        cin >> c;

    queue<array<int, 3>> Q;
    Q.push({y1, x1, 0});
    const constexpr int INF = 1e9;
    vector<vector<int>> D(h, vector<int>(w, INF));

    vector<int> dy = {0, 0, 1, -1}, dx = {1, -1, 0, 0};
    D[y1][x1] = 0;

    while (!Q.empty()){
        auto [y, x, d] = Q.front();
        Q.pop();
        for (int i = 0; i < 4; i++){
            for (int j = 1; j <= k; j++){
                int ny = y + dy[i] * j;
                int nx = x + dx[i] * j;
                if (ny < 0 or ny >= h or nx < 0 or nx >= w)
                    break;
                if (D[ny][nx] <= d)
                    break;
                if (C[ny][nx] == '@')
                    break;
              	if (D[ny][nx] <= d + 1)
                  	continue;
                Q.push({ny, nx, d + 1});
                D[ny][nx] = d + 1;
            }
        }
    }

    if (D[y2][x2] == INF){
        puts("-1");
        return 0;
    }

    cout << D[y2][x2] << endl;
}