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
#include <bits/stdc++.h>
using namespace std;

int main(){
    int H, W;
    cin >> H >> W;
    string A[1000];
    int dx[4] = {0, -1, 1, 0}, dy[4] = {1, 0, 0, -1};

    for(int i=0;i<H;i++) cin >> A[i];

    queue<pair<int,pair<int,int>>> Q;
    int ans = 0;
    for(int i=0;i<H;i++){
        for(int j=0;j<W;j++){
            if(A[i][j] == '#'){
                Q.push({0, {i, j}});
            }
        }
    }


    while(!Q.empty()){
        auto tp = Q.front();
        Q.pop();
        for(int i=0;i<4;i++){
            int x = tp.second.first + dx[i], y = tp.second.second + dy[i];
            if(x < 0 || x >= H || y < 0 || y >= W || A[x][y] == '#') continue;
            Q.push({tp.first+1, {x, y}});
            ans = max(ans, tp.first+1);
            A[x][y] = '#';
        }
    }

    cout << ans << endl;
    
    return 0;
}