Problem: You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black. HW characters from A_{11} to A_{HW} represent the colors of the squares. A_{ij} is `#` if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is `.` if that square is white.

We will repeatedly perform the following operation until all the squares are black:

* Every white square that shares a side with a black square, becomes black.



Find the number of operations that will be performed. The initial grid has at least one black square.

Constraints

* 1 \leq H,W \leq 1000
* A_{ij} is `#` or `.`.
* The given grid has at least one black square.

Input

Input is given from Standard Input in the following format:


H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}


Output

Print the number of operations that will be performed.

Examples

Input

3 3
...
.#.
...


Output

2


Input

6 6
..#..#
......
..#..
......
.#....
....#.


Output

3

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
typedef long long LL;
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;

int main(void)
{
  int h,w;
  cin >> h >> w;
  vector<string> f(h);
  for(auto &s:f) cin >> s;
  queue<pair<int,int> > q;
  REP(i,h) REP(j,w) {
    if(f[i][j]=='#') q.emplace(i,j);
    f[i][j]='.';
  }
  q.emplace(-1,-1);
  int answer = 0;
  int tmp = 0;
  while(q.size()>1) {
    int y,x;
    tie(y,x) = q.front();
    q.pop();
    if(y < 0) {
      tmp = -y;
      q.emplace(y-1,y-1);
      continue;
    } else if(f[y][x]!='#') {
      f[y][x]='#';
      if(y>0) q.emplace(y-1,x);
      if(y<h-1) q.emplace(y+1,x);
      if(x>0) q.emplace(y,x-1);
      if(x<w-1) q.emplace(y,x+1);
      answer = tmp;
    }
  }
  cout << answer << endl;
}