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)

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <functional>
#include <map>
#include <set>
#include <cmath>
#include <string>
#define SIZE 100005
#define INF 1000000000000005LL
#define ALL 1000000000000000LL

using namespace std;
typedef long long int ll;
typedef pair <ll,ll> P;

ll n;
int N;
ll K;
struct edge{ll to,cost;};
vector<edge> G[2][SIZE];//1:円 2:す
ll d[2][SIZE];

void dijkstra(ll s,int c){
	priority_queue<P, vector<P>, greater<P>> que;
	fill(d[c],d[c]+n,INF);
	d[c][s] = 0;
	que.push(P(0,s));

	while(!que.empty()){
		P p = que.top();que.pop();
		ll v = p.second;
		if(d[c][v] < p.first)continue;
		for(ll i=0;i<G[c][v].size();i++){
			edge e = G[c][v][i];
			if(d[c][e.to] > d[c][v] + e.cost){
				d[c][e.to] = d[c][v] + e.cost;
				que.push(P(d[c][e.to],e.to));
			}
		}
	}
}
ll dat[1<<19];

//配列、木:0オリジン
//n_:元の数列の要素数
//n:修正後の配列要素数 木のサイズ:2*n-1
void init(int n_){
	N=1;
	while(N<n_)N*=2;
	for(int i=0;i<2*N-1;i++)dat[i]=INF;
}

void update(int i,ll x){
	i += N-1;
	dat[i] = x;
	while(i>0){
		i = (i-1)/2;
		dat[i] = min(dat[i*2+1],dat[i*2+2]);
	}
}

//[a,b)のmin 節点k[l,r)について調べる
ll find(int a,int b,int k,int l,int r){
	if(r<=a || b<=l){
		return INF;
	}else if(a<=l && r<=b){
		return dat[k];
	}else{
		ll v1 = find(a,b,2*k+1,l,(l+r)/2);
		ll v2 = find(a,b,2*k+2,(l+r)/2,r);
		return min(v1,v2);
	}
}

int main()
{
	ll s,t;
	int m;
	scanf("%lld%d%lld%lld",&n,&m,&s,&t);
	ll u,v;
	ll a,b;
	for(int i=0;i<m;i++){
		scanf("%lld%lld%lld%lld",&u,&v,&a,&b);
		G[0][u].push_back((edge){v,a});
		G[0][v].push_back((edge){u,a});
		G[1][u].push_back((edge){v,b});
		G[1][v].push_back((edge){u,b});
	}
	n++;
	dijkstra(s,0);
	dijkstra(t,1);
	//d[0][i]+d[1][i] : コスト (1:n)
	N = (int)n;
	init(N);
	for(int i=1;i<=(int)(n-1);i++){
		//printf("%lld,",d[0][i]+d[1][i]);
		update(i-1,d[0][i]+d[1][i]);
	}
	ll cost = find(0,N,0,0,N);
	printf("%lld\n",ALL-cost);
	for(int i=0;i<(int)(n-2);i++){
		update(i,INF);
		ll cost = find(0,N,0,0,N);
		printf("%lld\n",ALL-cost);
	}

	return 0;
}