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 <iostream>
#include <iomanip>
#include <ios> 
#include <vector>
#include <string>
#include <algorithm>
#include <functional>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <bitset>
#include <map>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define sz(c) ((int)(c).size())

using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
const int MAX=1e5+5;
const ll INF=1e15;

struct edge{int to; ll cost;};
typedef pair<ll,int> P;
int N;
vector<edge> Gy[MAX], Gs[MAX];
ll dy[MAX],ds[MAX];

void dij_y(int s){
	priority_queue<P,vector<P>, greater<P>> quey;
	fill(dy+1, dy+N+1, INF);
	dy[s] = 0;
	quey.push(make_pair(0,s));
	
	while(!quey.empty()){
		P p = quey.top();quey.pop();
		int v=p.second;
		if(dy[v] < p.first) continue;
		for(int i = 0; i < Gy[v].size(); i++){
			edge e = Gy[v][i];
			if(dy[e.to] > dy[v]+e.cost){
				dy[e.to] = dy[v]+e.cost;
				quey.push(make_pair(dy[e.to], e.to));
			}
		}
	}
}

void dij_s(int s){
	priority_queue<P,vector<P>, greater<P>> ques;
	fill(ds+1, ds+N+1, INF);
	ds[s] = 0;
	ques.push(make_pair(0,s));
	
	while(!ques.empty()){
		P p = ques.top();ques.pop();
		int v=p.second;
		if(ds[v] < p.first) continue;
		for(int i = 0; i < Gs[v].size(); i++){
			edge e = Gs[v][i];
			if(ds[e.to] > ds[v]+e.cost){
				ds[e.to] = ds[v]+e.cost;
				ques.push(make_pair(ds[e.to], e.to));
			}
		}
	}
}

int main(){
	int M;
	cin>>N>>M;
	int s,t;
	cin>>s>>t;
	
	int u[MAX],v[MAX];
	ll a[MAX],b[MAX];
	rep(i,M){
		int u,v;
		ll a,b;
		cin>>u>>v>>a>>b;
		
		edge e={v,a};
		Gy[u].push_back(e);
		e={u,a};
		Gy[v].push_back(e);
		
		e={v,b};
		Gs[u].push_back(e);
		e={u,b};
		Gs[v].push_back(e);
	}
	dij_y(s);
	dij_s(t);
	
	ll d[MAX] ,dmin[MAX];
	for(int i=N;i>=1;i--){
		d[i]=dy[i]+ds[i];
		if(i==N)dmin[i]=d[i];
		else dmin[i]=min(d[i],dmin[i+1]);
	}
	rep1(i,N){
		cout<<INF-dmin[i]<<endl;
	}
}