#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC target("sse4,avx2,fma,avx")
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;
const int INF = 250000;
int n, h, w;
char g[500][500];
pair<int, int> end_pos[500][500][4];
int dp[500][500][9][9];
vector<pair<int, int>> q[INF];
inline bool inside(int x, int y) {
return (x >= 0 && y >= 0 && x < h && y < w && g[x][y] != 'x');
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n >> w >> h;
memset(dp, 0x3f, sizeof(dp));
FOR(i, 0, h) FOR(j, 0, w) {
cin >> g[i][j];
if (g[i][j] - '0' > 0 && g[i][j] - '0' < 10) {
dp[i][j][g[i][j] - '1'][g[i][j] - '1'] = 0;
}
FOR(k, 0, 4) end_pos[i][j][k] = {-1, -1};
}
// Determine final positions from each push
FOR(i, 0, h) FOR(j, 0, w) if (g[i][j] != 'x') {
FOR(k, 0, 4) {
pair<int, int> pos = {i, j};
int d = (k + (g[i][j] == 'A') * 3 + (g[i][j] == 'C')) % 4; // NESW
while (true) {
if (d == 0) {
if (inside(pos.first - 1, pos.second)) pos.first--;
else break;
} else if (d == 1) {
if (inside(pos.first, pos.second + 1)) pos.second++;
else break;
} else if (d == 2) {
if (inside(pos.first + 1, pos.second)) pos.first++;
else break;
} else {
if (inside(pos.first, pos.second - 1)) pos.second--;
else break;
}
if (~end_pos[pos.first][pos.second][d].first) {
pos = end_pos[pos.first][pos.second][d];
break;
}
if (g[pos.first][pos.second] == 'A') d = (d + 3) % 4;
if (g[pos.first][pos.second] == 'C') d = (d + 1) % 4;
}
end_pos[i][j][k] = pos;
}
}
// Find the minimum no. of pushes to get robot k-l to block (i, j)
FOR(rng, 0, n) {
FOR(k, 0, n - rng) {
int l = k + rng;
FOR(i, 0, h) FOR(j, 0, w) if (g[i][j] != 'x') {
FOR(d, k, l) {
dp[i][j][k][l] =
min(dp[i][j][k][l],
dp[i][j][k][d] + dp[i][j][d + 1][l]);
}
}
}
FOR(k, 0, n - rng) {
int l = k + rng;
FOR(i, 0, h) FOR(j, 0, w) if (g[i][j] != 'x') {
if (dp[i][j][k][l] <= INF) q[dp[i][j][k][l]].push_back({i, j});
}
FOR(i, 0, INF) {
for (pair<int, int> pos : q[i]) {
int x, y;
tie(x, y) = pos;
if (dp[x][y][k][l] == i) {
FOR(d, 0, 4) {
int nx, ny;
tie(nx, ny) = end_pos[x][y][d];
if (dp[nx][ny][k][l] > dp[x][y][k][l] + 1) {
q[dp[nx][ny][k][l] = dp[x][y][k][l] + 1].push_back({nx, ny});
}
}
}
}
q[i].clear();
}
}
}
int ans = INT_MAX;
FOR(i, 0, h) FOR(j, 0, w) ans = min(ans, dp[i][j][0][n - 1]);
cout << (ans > INF ? -1 : ans);
return 0;
}
