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<bits/stdc++.h>

using namespace std;

#define LL long long
#define N 300000
#define P(x,y) make_pair(x,y)
#define ss second

const LL sum=1000000000000000LL;
LL n,m,s,t,x,y,a,b,tot,f[N],g[N],head[N],ans[N];
bool vis[N];
priority_queue<pair<LL,LL> > pq;
struct edge{LL v,a,b,nxt;}e[N];

void add(LL x,LL y,LL a,LL b){
	e[++tot].v=y; e[tot].a=a; e[tot].b=b; e[tot].nxt=head[x]; head[x]=tot;
}

int main(){
	scanf("%lld%lld%lld%lld",&n,&m,&s,&t);
	for (LL i=1;i<=m;++i){
		scanf("%lld%lld%lld%lld",&x,&y,&a,&b);
		add(x,y,a,b); add(y,x,a,b);
	}
	memset(f,0x3f,sizeof f); f[s]=0;
	memset(vis,0,sizeof vis);
	pq.push(P(-f[s],s));
	while (!pq.empty()){
		LL u; for (u=pq.top().ss,pq.pop();vis[u]&&!pq.empty();u=pq.top().ss,pq.pop());
		if (vis[u]) break; vis[u]=1;
		for (LL i=head[u],v;i;i=e[i].nxt)
			if (f[v=e[i].v]>f[u]+e[i].a){
				f[v]=f[u]+e[i].a;
				pq.push(P(-f[v],v));
			}
	}
	memset(g,0x3f,sizeof g); g[t]=0;
	memset(vis,0,sizeof vis);
	pq.push(P(-g[t],t));
	while (!pq.empty()){
		LL u; for (u=pq.top().ss,pq.pop();vis[u]&&!pq.empty();u=pq.top().ss,pq.pop());
		if (vis[u]) break; vis[u]=1;
		for (LL i=head[u],v;i;i=e[i].nxt)
			if (g[v=e[i].v]>g[u]+e[i].b){
				g[v]=g[u]+e[i].b;
				pq.push(P(-g[v],v));
			}
	}
	for (LL i=n;i;--i)
		ans[i]=max(ans[i+1],sum-f[i]-g[i]);
	for (LL i=1;i<=n;++i) printf("%lld\n",ans[i]);
	
	return 0;
}