#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>using namespace std;typedef unsigned long long ull;
const int maxn = 1e5+10;
ull x[maxn];
int ans[maxn];struct Edge {int lst;int to;
};class Tree {
public :Edge edge[maxn<<1];int head[maxn];int cn, csz, ccnt, mid[3];int rcd[maxn];ull cnode[maxn];ull vvvvv[3];map<string, int> id;char name[maxn][16];inline void add(int u, int v) {edge[csz].lst = head[u];edge[csz].to  = v;head[u] = csz++;}int dfs(int u, int fa) {int i, v, res = 0, t1;for (i=head[u]; i; i=edge[i].lst) {v = edge[i].to;if (v == fa) continue;t1 = dfs(v, u);res += t1;if (rcd[u] < t1) rcd[u] = t1;if (rcd[v] < cn-t1) rcd[v] = cn-t1; }return res + 1;}ull dfs2(int u, int fa, int deep) {int i, v, res = 0, t1;vector<ull> son;for (i=head[u]; i; i=edge[i].lst) {v = edge[i].to;if (v == fa) continue; son.push_back(dfs2(v, u, deep+1));}sort(son.begin(), son.end()); // 判断树同构 注意要sort后乘一个随机数(或者一个质数的i次方).  为什么可以sort呢? 因为树同一层的节点围绕父节点旋转不会改变树的结构. 而sort相当于把节点绕父节点旋转了. for (i=0, t1=son.size(); i<t1; ++i) res += son[i] * x[i+1];return cnode[u] = (res ? res : x[deep]);}void init(int n) {cn = n; csz = 1; id.clear();int i, u, v, cnt = 1;char s1[16], s2[16];for (i=1; i<=n; ++i) rcd[i] = head[i] = 0;for (i=1; i<n; ++i) {scanf("%s%s", s1, s2);if (!(u = id[s1])) {strcpy(name[cnt], s1);id[s1] = u = cnt++;}if (!(v = id[s2])) {strcpy(name[cnt], s2);id[s2] = v = cnt++;}add(u, v); add(v, u);}dfs(1, -1);int mm = 0x3f3f3f3f;for (i=1; i<=n; ++i) {if (mm > rcd[i]) {mm = rcd[i];ccnt = 0;mid[++ccnt] = i;} else if (mm == rcd[i]) mid[++ccnt] = i;}for (i=1; i<=ccnt; ++i) vvvvv[i] = dfs2(mid[i], -1, 1);}
}te1, te2;struct nobe {int id;ull val;bool operator < (const nobe &a) const {return val < a.val;}nobe () {}nobe (int iid, ull vval) : id(iid), val(vval) {}
};void dfs3(int u1, int u2, int fa1, int fa2) {int i, v1, v2, tsz;ans[u1] = u2;vector<nobe> ve1, ve2;for (i=te1.head[u1]; i; i=te1.edge[i].lst) {v1 = te1.edge[i].to;if (v1 == fa1) continue;ve1.push_back(nobe(v1, te1.cnode[v1]));}for (i=te2.head[u2]; i; i=te2.edge[i].lst) {v2 = te2.edge[i].to;if (v2 == fa2) continue;ve2.push_back(nobe(v2, te2.cnode[v2]));}sort(ve1.begin(), ve1.end());sort(ve2.begin(), ve2.end());for (i=0, tsz=ve1.size(); i<tsz; ++i) dfs3(ve1[i].id, ve2[i].id, u1, u2);
}int main()
{int i, j, k, n;srand(time(NULL));for (i=0; i<maxn; ++i) x[i] = rand();while (~scanf("%d", &n)) {te1.init(n);te2.init(n);for (i=1; i<=te1.ccnt; ++i) for (j=1; j<=te2.ccnt; ++j) {if (te1.vvvvv[i] == te2.vvvvv[j]) {dfs3(te1.mid[i], te2.mid[j], te1.mid[i], te2.mid[j]);for (k=1; k<=n; ++k) printf("%s %s\n", te1.name[k], te2.name[ans[k]]);goto A;}}A: ;}return 0;
}