## Algorithm

E1. Stars Drawing (Easy Edition)
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

star is a figure of the following type: an asterisk character '*' in the center of the figure and four rays (to the left, right, top, bottom) of the same positive length. The size of a star is the length of its rays. The size of a star must be a positive number (i.e. rays of length 00 are not allowed).

Let's consider empty cells are denoted by '.', then the following figures are stars:

The leftmost figure is a star of size 11, the middle figure is a star of size 22 and the rightmost figure is a star of size 33.

You are given a rectangular grid of size n×m�×� consisting only of asterisks '*' and periods (dots) '.'. Rows are numbered from 11 to n, columns are numbered from 11 to m. Your task is to draw this grid using any number of stars or find out that it is impossible. Stars can intersect, overlap or even coincide with each other. The number of stars in the output can't exceed nm�⋅�. Each star should be completely inside the grid. You can use stars of same and arbitrary sizes.

In this problem, you do not need to minimize the number of stars. Just find any way to draw the given grid with at most nm�⋅� stars.

Input

The first line of the input contains two integers n and m (3n,m1003≤�,�≤100) — the sizes of the given grid.

The next n lines contains m characters each, the i-th line describes the i-th row of the grid. It is guaranteed that grid consists of characters '*' and '.' only.

Output

If it is impossible to draw the given grid using stars only, print "-1".

Otherwise in the first line print one integer k (0knm0≤�≤�⋅�) — the number of stars needed to draw the given grid. The next k lines should contain three integers each — xj��yj�� and sj��, where xj�� is the row index of the central star character, yj�� is the column index of the central star character and sj�� is the size of the star. Each star should be completely inside the grid.

Examples
input
Copy
6 8....*......**.....*****....**.......*...........
output
Copy
33 4 13 5 23 5 1
input
Copy
5 5.*...****..****..**......
output
Copy
32 2 13 3 13 4 1
input
Copy
5 5.*...***...*....*........
output
Copy
-1
input
Copy
3 3*.*.*.*.*
output
Copy
-1
Note

In the first example the output

23 4 13 5 2

is also correct.

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

struct node {
int x, y, c;
node() {}
node(int x, int y, int c) :
x(x), y(y), c(c) {}
};

int n, m, ii[4], jj[4];
bool vis[101][101];
char g[101][101];
vector<node> sol;

bool check() {
for(int i = 0; i < 4; ++i)
if(ii[i] < 0 || ii[i] >= n || jj[i] < 0 || jj[i] >= m || g[ii[i]][jj[i]] != '*')
return false;
return true;
}

int can(int i, int j) {
int c = 0;
if(g[i][j] != '*')
return c;
ii[0] = i + 1, jj[0] = j;
ii[1] = i - 1, jj[1] = j;
ii[2] = i, jj[2] = j + 1;
ii[3] = i, jj[3] = j - 1;
while(check()) {
++c;
for(int k = 0; k < 4; ++k)
vis[ii[k]][jj[k]] = true;
++ii[0];
--ii[1];
++jj[2];
--jj[3];
}
if(c > 0)
vis[i][j] = true;
return c;
}

int main() {
scanf("%d %d", &n, &m);
for(int i = 0; i < n; ++i)
scanf("%s", g[i]);

for(int i = 0; i < n; ++i)
for(int j = 0, cur; j < m; ++j) {
cur = can(i, j);
if(cur > 0)
sol.push_back(node(i, j, cur));
}

for(int i = 0; i < n; ++i)
for(int j = 0; j < m; ++j)
if(g[i][j] == '*' && !vis[i][j]) {
puts("-1");
return 0;
}

printf("%d\n", int(sol.size()));
for(int i = 0; i < sol.size(); ++i)
printf("%d %d %d\n", sol[i].x + 1, sol[i].y + 1, sol[i].c);

return 0;
}
Copy The Code &

Input

cmd
6 8
....*...
...**...
..*****.
...**...
....*...
........

Output

cmd
3
3 4 1
3 5 2
3 5 1
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## Demonstration

Codeforces Solution-E1. Stars Drawing (Easy Edition)-Solution in C, C++, Java, Python