#include <iostream>
#include <iomanip>
#include <string>
#include <stack>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <regex>
using namespace std;
using pii  = pair<int,int>;
using ll=long long;
using ld=long double;
#define pb push_back
#define mp make_pair
#define sc second
#define fr first
#define stpr setprecision
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define rep(i,n) for(ll i=0;i<(n);++i)
#define Rep(i,a,b) for(ll i=(a);i<(b);++i)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rRep(i,a,b) for(int i=a;i>=b;i--)
#define crep(i) for(char i='a';i<='z';++i)
#define psortsecond(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.second<b.second;});
#define ALL(x) (x).begin(),(x).end()
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define endl '\n'
int ctoi(const char c){
  if('0' <= c && c <= '9') return (c-'0');
  return -1;
}
ll gcd(ll a,ll b){return (b == 0 ? a : gcd(b, a%b));}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
constexpr ll MOD=1000000007;
constexpr ll INF=1000000011;
constexpr ll MOD2=998244353;
constexpr ll LINF = 1001002003004005006ll;
constexpr ld EPS=10e-8;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> istream& operator>>(istream& is,vector<T>& v){for(auto&& x:v)is >> x;return is;}
template<typename T,typename U> istream& operator>>(istream& is, pair<T,U>& p){ is >> p.first; is >> p.second; return is;}
template<typename T,typename U> ostream& operator>>(ostream& os, const pair<T,U>& p){ os << p.first << ' ' << p.second; return os;}
template<class T> ostream& operator<<(ostream& os, vector<T>& v){
    for(auto i=begin(v); i != end(v); ++i){
        if(i !=begin(v)) os << ' ';
        os << *i;
    }
    return os;
}

ll N,M;
vector<pair<ll,ll>> A[100007],B[100007];   //A[i].firstはA[i]枝の先　A[i].secondは重さを表す
ll DKSTR1[100007],DKSTR2[100007];   //距離の最小値を記入する
ll chek[100007];

ll dijkstra1(ll s,ll t,ll g){   // s始点　t初期値の重さ　g目的地　sからtへの最短距離を返す
    Rep(i,0,N+1){
        chek[i]=0;
        DKSTR1[i]=LINF;
    }
    DKSTR1[s]=t;
    ll key=s;
    priority_queue <pair<ll,ll>> PQ;
    PQ.push(mp(t,s));
    while(PQ.empty()!=1){
        pair<ll,ll> f = PQ.top();
        PQ.pop();
        key=f.second;
        chek[key]=1;
        if(DKSTR1[key] < f.first*(-1)){
            continue;
        }
        rep(j,A[key].size()){
            ll v = A[key][j].first;
            if(chek[v]==0 && DKSTR1[v]>DKSTR1[key]+A[key][j].second){
                DKSTR1[v] = DKSTR1[key] + A[key][j].second;
                PQ.push(mp(DKSTR1[v]*(-1),v));
            }
        }
    }
    return DKSTR1[g];
}

ll dijkstra2(ll s,ll t,ll g){   // s始点　t初期値の重さ　g目的地　sからtへの最短距離を返す
    Rep(i,0,N+1){
        chek[i]=0;
        DKSTR2[i]=LINF;
    }
    DKSTR2[s]=t;
    ll key=s;
    priority_queue <pair<ll,ll>> PQ;
    PQ.push(mp(t,s));
    while(PQ.empty()!=1){
        pair<ll,ll> f = PQ.top();
        PQ.pop();
        key=f.second;
        chek[key]=1;
        if(DKSTR2[key] < f.first*(-1)){
            continue;
        }
        rep(j,B[key].size()){
            ll v = B[key][j].first;
            if(chek[v]==0 && DKSTR2[v]>DKSTR2[key]+B[key][j].second){
                DKSTR2[v] = DKSTR2[key] + B[key][j].second;
                PQ.push(mp(DKSTR2[v]*(-1),v));
            }
        }
    }
    return DKSTR2[g];
}

int main(){    //この問題は同じ道を通らず自分に戻ってくる最短経路を求める問題
    ll X,Y,L1,L2,S,T,mi = 0;
    cin >> N >> M >> S >> T;
    rep(i,M){
        cin >> X >> Y >> L1 >> L2;   // X=始点　Y=終点  L=重さ
        A[X].pb(mp(Y,L1));
        A[Y].pb(mp(X,L1));
        B[X].pb(mp(Y,L2));
        B[Y].pb(mp(X,L2));
    }
    ll K=dijkstra1(S,0,T);
    K=dijkstra2(T,0,S);
    ll  C[100007],D[100007];
    rRep(i,N,1){
      // cout << DKSTR1[i] << " " << DKSTR2[i] << endl;
      C[i] = DKSTR1[i] + DKSTR2[i];
      mi = max(mi,1000000000000000-C[i]);
      D[i] = mi;
    }
    Rep(i,1,N+1){
      cout << D[i] << endl;
    }
}

Measured memory_bytes: 19908