Problem:
Score : 400 points

Problem Statement

Kenkoooo is planning a trip in Republic of Snuke.
In this country, there are n cities and m trains running.
The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally.
Any city can be reached from any city by changing trains.

Two currencies are used in the country: yen and snuuk.
Any train fare can be paid by both yen and snuuk.
The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.

In a city with a money exchange office, you can change 1 yen into 1 snuuk.
However, when you do a money exchange, you have to change all your yen into snuuk.
That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk.
Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.

Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling.
It is acceptable to do the exchange in City s or City t.

Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange.
For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years.
You can assume that the trip finishes within the year.

Constraints

2 \leq n \leq 10^5

1 \leq m \leq 10^5

1 \leq s,t \leq n

s \neq t

1 \leq u_i < v_i \leq n

1 \leq a_i,b_i \leq 10^9

If i\neq j, then u_i \neq u_j or v_i \neq v_j.

Any city can be reached from any city by changing trains.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m

Output

Print n lines.
In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/* --- author: dxm --- */
#include<bits/stdc++.h>
using namespace std;

#define INF 1e9
#define Inf 1e18
#define rep(i,n) for(int (i)=0;(i)<n;i++)
#define REP(i,n) for(int (i)=1;(i)<=n;i++)
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define fst first
#define snd second
typedef long long ll;
typedef double db;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
#define DEBUG

const int maxn=100005;
int n,m,s,t;
vector<pii>edge[maxn];
ll dist[maxn][2],a[maxn],b[maxn],ans[maxn];
priority_queue<pair<ll,int> >q;

int main(){
#ifndef DEBUG
	freopen("my.in","r",stdin);
	freopen("my.out","w",stdout);
#endif
	scanf("%d%d%d%d",&n,&m,&s,&t);
	REP(i,m){
		int u,v;
		scanf("%d%d%d%d",&u,&v,&a[i],&b[i]);
		edge[u].pb(mp(v,i));
		edge[v].pb(mp(u,i));
	}
	REP(i,n)rep(j,2)dist[i][j]=Inf;
	dist[s][0]=0;
	q.push(mp(0,s));
	while(!q.empty()){
		pair<ll,int> x=q.top();q.pop();
		x.fst=-x.fst;
		if(dist[x.snd][0]!=x.fst)continue;
		rep(i,edge[x.snd].size()){
			pii y=edge[x.snd][i];
			if(dist[y.fst][0]>x.fst+a[y.snd]){
				dist[y.fst][0]=x.fst+a[y.snd];
				q.push(mp(-dist[y.fst][0],y.fst));
			}
		}
	}
	dist[t][1]=0;
	q.push(mp(0,t));
	while(!q.empty()){
		pair<ll,int> x=q.top();q.pop();
		x.fst=-x.fst;
		if(dist[x.snd][1]!=x.fst)continue;
		rep(i,edge[x.snd].size()){
			pii y=edge[x.snd][i];
			if(dist[y.fst][1]>x.fst+b[y.snd]){
				dist[y.fst][1]=x.fst+b[y.snd];
				q.push(mp(-dist[y.fst][1],y.fst));
			}
		}
	}
	ans[n+1]=Inf;
	for(int i=n;i>=1;i--){
		ans[i]=ans[i+1];
		ans[i]=min(ans[i],dist[i][0]+dist[i][1]);
	}
	REP(i,n)printf("%lld\n",1LL*100000*100000*100000-ans[i]);
	return 0;
}

/*
Input:
-----------------
Output:
*/