#include #define rep(i, a) for (int i = 0; i < (a); i++) #define rep2(i, a, b) for (int i = (a); i < (b); i++) using namespace std; typedef long long ll; const ll inf = 1e9; const ll mod = 1e9 + 7; int N, A[50]; class Dinic { public: struct edge { int to, cap, rev; }; vector > G; Dinic() {} Dinic(int V) { resize(V); } void resize(int V) { G.resize(V); level.resize(V); iter.resize(V); } void add(int u, int v, int cap) { G[u].push_back((edge){v, cap, (int)G[v].size()}); G[v].push_back((edge){u, 0, (int)G[u].size() - 1}); } int calc(int s, int t) { int res = 0; for (;;) { bfs(s); if (level[t] < 0) return res; fill(iter.begin(), iter.end(), 0); int f; while ((f = dfs(s, t, inf)) > 0) { res += f; } } } private: vector level; vector iter; void bfs(int s) { fill(level.begin(), level.end(), -1); queue q; q.push(s); level[s] = 0; while (!q.empty()) { int v = q.front(); q.pop(); for (int i = 0; i < G[v].size(); i++) { edge &e = G[v][i]; if (e.cap == 0) continue; if (level[e.to] >= 0) continue; level[e.to] = level[v] + 1; q.push(e.to); } } } int dfs(int v, int t, int f) { if (v == t) return f; for (int &i = iter[v]; i < G[v].size(); i++) { edge &e = G[v][i]; if (e.cap == 0) continue; if (level[v] >= level[e.to]) continue; int flow = dfs(e.to, t, min(f, e.cap)); if (flow == 0) continue; e.cap -= flow; G[e.to][e.rev].cap += flow; return flow; } return 0; } }; class BipartiteMatching { public: Dinic dinic; BipartiteMatching(int U, int V) : U(U), V(V) { dinic.resize(U + V + 2); rep (i, U) dinic.add(U + V, i, 1); rep (j, V) dinic.add(j + U, U + V + 1, 1); } void add(int u, int v) { dinic.add(u, U + v, 1); } int calc() { return dinic.calc(U + V, U + V + 1); } private: int U, V; }; int main() { cin >> N; rep (i, N) cin >> A[i]; BipartiteMatching m(N, N); rep (i, N) { rep (j, N) if (j != A[i]) { m.add(i, j); } } m.calc(); vector ans; bool ok = true; rep (i, N) { for (auto e : m.dinic.G[i]) { if (e.cap == 0 && e.to - N < N) { ans.push_back(e.to - N); } } } if (ans.size() == N) { for (auto x : ans) { cout << x << endl; } } else { cout << -1 << endl; } return 0; }