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)

//00:10
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
#define ll long long
const ll INF=1e18;
ll DA[N],DB[N];
int Ord[N];
bool cmp(int a,int b)
{
	return DA[a]+DB[a]<DA[b]+DB[b];
}
struct Edge
{
	int v,w[2];
};
vector<Edge>V[N];
struct T
{
	int id;ll sum;
	bool operator<(T b)const
	{
		return sum>b.sum;
	}
};
priority_queue<T>Q;
int n,m;
void Dij(ll*sp,int s,int flag)
{
	for(int i=1;i<=n;i++)sp[i]=INF;
	sp[s]=0;Q.push({s,0});
	while(!Q.empty()){
		T p=Q.top();Q.pop();
		if(p.sum>sp[p.id])continue;
		for(auto k:V[p.id]){
			if(sp[p.id]+k.w[flag]<sp[k.v]){
				sp[k.v]=sp[p.id]+k.w[flag];
				Q.push({k.v,sp[k.v]});
			}
		}
	}
}
bool vis[N];
const ll Cost=1e15;
int main()
{
	//freopen("/home/zz7/CFInput","r",stdin);
	int s,t;scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=0;i<m;i++){
		int u,v,a,b;
		scanf("%d%d%d%d",&u,&v,&a,&b);
		V[u].push_back({v,a,b});
		V[v].push_back({u,a,b});
	}
	Dij(DA,s,0);
	Dij(DB,t,1);
	for(int i=0;i<n;i++)Ord[i]=i+1;
	sort(Ord,Ord+n,cmp);
	int j=0;
	for(int i=0;i<n;i++){
		vis[i]=true;
		while(vis[Ord[j]])j++;
		cout<<Cost-(DA[Ord[j]]+DB[Ord[j]])<<'\n';
	}
}