#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;
typedef long long ll;

class Point {
public:
    int x, y;
    int step;
    int status;
    Point(int i, int j, int t, int s) : x(i), y(j), step(t), status(s) {}
    Point() {}
    friend bool operator==(const Point &A, const Point &B) { return (A.x == B.x && A.y == B.y); }
};

int bfs(Point st, int k);
bool is_passible(int x, int y);

const int dx[] = { 1, -1, 0, 0 };
const int dy[] = { 0, 0, 1, -1 };
const int MAX = 35, MAXST = (1 << 10) + 5;
char graph[MAX][MAX];
bool visited[MAX][MAX][MAXST];
int n, m;

int main() {
    // freopen("test.txt", "r", stdin);
    while (scanf("%d %d", &n, &m) != EOF) {
        // 初始化
        memset(visited, false, sizeof(visited));
        // 输入
        for (int i = 1; i <= n; i++) {
            scanf("%s", graph[i] + 1);
        }
        // 预处理
        int k = 0;
        Point st;
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++) {
                if (graph[i][j] == 'p') {
                    st = Point(i, j, 0, 0);
                }
                if (graph[i][j] == '@') {
                    graph[i][j] = k + '0';
                    k++;
                }
            }
        // BFS
        int res = bfs(st, k);
        // 输出
        printf("%d\n", res);
    }
    return 0;
}
// 广度优先搜索
int bfs(Point st, int k) {
    // 主循环
    queue<Point> q;
    q.push(st);
    while (!q.empty()) {
        // 出队
        Point p = q.front();
        q.pop();
        // 遍历
        for (int i = 0; i < 4; i++) {
            // 获取目标坐标
            int nx = p.x + dx[i], ny = p.y + dy[i];
            int nstep = p.step + 1;
            int nstatus = p.status;
            // 不可通行的情况
            if (!is_passible(nx, ny)) {
                continue;
            }
            // 更新状态
            if (graph[nx][ny] >= '0' && graph[nx][ny] <= '9') {
                nstatus |= (1 << (graph[nx][ny] - '0'));
            }
            // 完成的情况
            if (nstatus == (1 << k) - 1) {
                return nstep;
            }
            // 入队
            if (!visited[nx][ny][nstatus]) {
                visited[nx][ny][nstatus] = true;
                q.push(Point(nx, ny, nstep, nstatus));
            }
        }
    }
    // 无解的情况
    return -1;
}
// 判断是否可以通行
inline bool is_passible(int x, int y) { return (graph[x][y] != '#' && x >= 1 && x <= n && y >= 1 && y <= m); }