Problem: There are some coins in the xy-plane. The positions of the coins are represented by a grid of characters with H rows and W columns. If the character at the i-th row and j-th column, s_{ij}, is `#`, there is one coin at point (i,j); if that character is `.`, there is no coin at point (i,j). There are no other coins in the xy-plane.

There is no coin at point (x,y) where 1\leq i\leq H,1\leq j\leq W does not hold. There is also no coin at point (x,y) where x or y (or both) is not an integer. Additionally, two or more coins never exist at the same point.

Find the number of triples of different coins that satisfy the following condition:

* Choosing any two of the three coins would result in the same Manhattan distance between the points where they exist.



Here, the Manhattan distance between points (x,y) and (x',y') is |x-x'|+|y-y'|. Two triples are considered the same if the only difference between them is the order of the coins.

Constraints

* 1 \leq H,W \leq 300
* s_{ij} is `#` or `.`.

Input

Input is given from Standard Input in the following format:


H W
s_{11}...s_{1W}
:
s_{H1}...s_{HW}


Output

Print the number of triples that satisfy the condition.

Examples

Input

5 4
#.##
.##.
#...
..##
...#


Output

3


Input

5 4
.##
.##.
...
..##
...#


Output

3


Input

13 27
......#.........#.......#..
...#.....###..
..............#####...##...
...#######......#...#######
...#.....#.....###...#...#.
...#######....#.#.#.#.###.#
..............#.#.#...#.#..
.#.#.#...###..
...........#...#...#######
..#######..#...#...#.....#
..#.....#..#...#...#.###.#
..#######..#...#...#.#.#.#
..........##...#...#.#####


Output

870

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 305;
int n, m;
char mp[3*N][3*N];
int suml[3*N][3*N], sumr[3*N][3*N];
ll solve(int x, int y) {
    ll ans = 0;
    for (int i = 1; i <= n; i++) {
        int num = 0;
        if (mp[x-i][y] == '#') num++;
        if (mp[x+i][y] == '#') num++;
        if (mp[x][y-i] == '#') num++;
        if (mp[x][y+i] == '#') num++;
        if (num == 3) ans++;
        else if (num == 4) ans += 4;
    }
    for (int i = 2; x-i >= N+1 && y-i >= N+1; i++) {
        if (mp[x-i][y] == '.' || mp[x][y-i] == '.') continue;
        int sx = x+1, sy = y+i-1;
        int ex = x+i-1, ey = y+1;
        ans += suml[ex][ey] - suml[sx-1][sy+1];
    }
    for (int i = 2; x-i >= N+1 && y+i <= N+m; i++) {
        if (mp[x-i][y] == '.' || mp[x][y+i] == '.') continue;
        int sx = x+1, sy = y-i+1;
        int ex = x+i-1, ey = y-1;
        ans += sumr[ex][ey] - sumr[sx-1][sy-1];
    }
    for (int i = 2; x+i <= N+n && y+i <= N+m; i++) {
        if (mp[x+i][y] == '.' || mp[x][y+i] == '.') continue;
        int sx = x-i+1, sy = y-1;
        int ex = x-1, ey = y-i+1;
        ans += suml[ex][ey] - suml[sx-1][sy+1];
    }
    for (int i = 2; x+i <= N+n && y-i >= N+1; i++) {
        if (mp[x+i][y] == '.' || mp[x][y-i] == '.') continue;
        int sx = x-i+1, sy = y+1;
        int ex = x-1, ey = y+i-1;
        ans += sumr[ex][ey] - sumr[sx-1][sy-1];
    }
    return ans;
}
int main() {
    scanf("%d%d", &n, &m);
    for (int i = N+1; i <= N+n; i++) scanf("%s", mp[i]+1+N);
    for (int i = 1; i <= 2*N+n; i++)
        for (int j = 1; j <= 2*N+m; j++)
            suml[i][j] = sumr[i][j] = (mp[i][j] == '#');
    for (int i = 1; i <= 2*N+n; i++)
        for (int j = 1; j <= 2*N+m; j++) {
            sumr[i][j] += sumr[i-1][j-1];
            suml[i][j] += suml[i-1][j+1];
        }
    ll ans = 0;
    for (int i = N+1; i <= N+n; i++)
        for (int j = N+1; j <= N+m; j++)
            ans += solve(i, j);
    cout << ans << endl;
}
/*
6 6
...#..
..#...
.#...#
......
...#..
......
*/