#include<atcoder/dsu>
#include<iostream>
#include<stdio.h>
#include<vector>
#include<deque>
#include<algorithm>
#include<utility>
using namespace atcoder;
using namespace std;
#define ll long long
#define P pair<ll,ll>
#define Pl pair<pair<ll,ll>,ll>

ostream& operator<<(ostream& os,const P& a){
  return os<<a.first<<" "<<a.second;
}

ostream& operator<<(ostream& os,const vector<ll>& a){
  for(ll i:a){
    os<<i<<"\n";
  }
  return os;
}

ostream& operator<<(ostream& os,const vector<vector<int>>& a){
  for(vector<int> i:a){
    for(int j:i){
      os<<j<<" ";
    }
    os<<"\n";
  }
  return os;
}

ostream& operator<<(ostream& os,const vector<P>& a){
  for(P i:a){
    os<<i<<"\n";
  }
  return os;
}

signed main(){
  ll N,A,B;
  cin>>N;
  dsu uf(N);
  vector<vector<P>>to(N);
  for(int i=0;i<N;i++){
    cin>>A>>B;
    uf.merge(A-1,B-1);
    to.at(A-1).push_back({B-1,i});
    if(A!=B){
      to.at(B-1).push_back({A-1,i});
    }
  }
  vector<ll> seen(N);
  vector<ll>ans(N,-1);
  vector<vector<int>> groups=uf.groups();
  for(vector<int> linked:groups){
    bool is_cycle=true;
    ll root=-1,cycle_cnt=0;
    for(int point:linked){
      ll link_size=to.at(point).size();
      for(ll k=0;k<link_size;k++){
        if(to.at(point).at(k).first==point){
          cycle_cnt++;
          is_cycle=false;
          ans.at(to.at(point).at(k).second)=point+1;
          root=point;
        }
      }
    }
    if(cycle_cnt>1){
      //自己ループが2つ以上
      printf("No\n");
      cerr<<"1\n";
      return 0;
    }
    if(is_cycle){
      ll next_point=-1;
      deque<P>que;
      que.push_back({linked.at(0),-1});
      while(!que.empty()){
        P q=que.back();
        que.pop_back();
        for(P j:to.at(q.first)){
          if(j.second==q.second)continue;
          if(seen.at(j.first)){
            root=j.first;
            next_point=q.first;
            ans.at(j.second)=root+1;
            que.clear();
            break;
          }
          que.push_back({j.first,j.second});
        }
        seen.at(q.first)=1;
      }
      if(next_point==-1){
        //頂点が1つのとき、かつ辺の数が0のとき
        printf("No\n");
        cerr<<"2\n";
        return 0;
      }
    }
    deque<ll>dq;
    dq.push_back(root);
    while(!dq.empty()){
      ll q=dq.back();dq.pop_back();
      for(P j:to.at(q)){
        if(ans.at(j.second)==-1){
          ans.at(j.second)=j.first+1;
          dq.push_back(j.first);
        }
      }
    }
  }
  for(ll i:ans){
    if(i==-1){
      //塗れていないとき
      printf("No\n");
      cerr<<"3\n";
      return 0;
    }
  }
  cout<<"Yes\n";
  cout<<ans;
  return 0;
}