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: Wrong Answer)



#include <bits/stdc++.h>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define REP(i,n) for(int i=0;i<n;i++)
#define FOR(i,n1,n2) for(int i=n1;i<n2;i++)
#define speed_up    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

typedef long long int ll;
typedef pair<ll,ll> Pi;
const int INF=(ll)(1LL<<31)-1;
const ll INFl=(ll)9223372036854775807;
const int MAX=10000;
const ll MOD=(ll)1e9+7;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a, b)*b;}
//int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

int n,m,s,t;
int u[100100],v[100100],a[100100],b[100100];
typedef struct{
	int to;
	int costy;
	int costs;
}edge;

ll dy[100100];
ll ds[100100];
vector<edge> G[100100];

int main(){
	cin>>n>>m>>s>>t;
	s--,t--;
	REP(i,m){
		cin>>u[i]>>v[i]>>a[i]>>b[i];
		u[i]--,v[i]--;
		edge e1={v[i],a[i],b[i]};
		edge e2={u[i],a[i],b[i]};
		G[u[i]].push_back(e1);
		G[v[i]].push_back(e2);
	}
	fill(dy,dy+n,INF);
	fill(ds,ds+n,INF);
	dy[s]=0;
	ds[t]=0;
	priority_queue<Pi,vector<Pi>,greater<Pi> > quey;
	priority_queue<Pi,vector<Pi>,greater<Pi> > ques;
	quey.push(Pi(0,s));
	ques.push(Pi(0,t));
	while(!quey.empty()){
		Pi p=quey.top();quey.pop();
		int v=p.second;
		if(dy[v]<p.first)continue;
		for(int i=0;i<G[v].size();i++){
			edge e=G[v][i];
			if(dy[e.to]>dy[v]+e.costy){
				dy[e.to]=dy[v]+e.costy;
				quey.push(Pi(dy[e.to],e.to));
			}
		}
	}
	
	while(!ques.empty()){
		Pi p=ques.top();ques.pop();
		int v=p.second;
		if(ds[v]<p.first)continue;
		for(int i=0;i<G[v].size();i++){
			edge e=G[v][i];
			if(ds[e.to]>ds[v]+e.costs){
				ds[e.to]=ds[v]+e.costs;
				ques.push(Pi(ds[e.to],e.to));
			}
		}
	}
	ll ccost[100100];
	ll ans[100100];
	for(int i=0;i<n;i++)
		ccost[i]=dy[i]+ds[i];
	ans[n-1]=ccost[n-1];
	for(int i=n-2;i>=0;i--){
		ans[i]=min(ccost[i],ans[i+1]);
	}
	for(int i=0;i<n;i++){
		cout<<1000000000000000-ans[i]<<endl;
	}
	return 0;
}