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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define REP(i,n) for (long long i=0;i<(n);i++)
#define FOR(i,a,b) for (long long i=(a);i<(b);i++)
#define RREP(i,n) for(long long i=n;i>=0;i--)
#define RFOR(i,a,b) for(long long i=(a);i>(b);i--)
#define dump1d_arr(array) REP(i,array.size()) cerr << #array << "[" << (i) << "] ==> " << (array[i]) << endl
#define dump2d_arr(array) REP(i,array.size()) REP(j,array[i].size()) cerr << #array << "[" << (i) << "]" << "[" << (j) << "] ==> " << (array[i][j]) << endl
#define dump(x)  cerr << #x << " => " << (x) << endl


#define dumpP(p) cerr << "( " << p.first << " , " << p.second << " )" << ends
#define SORT(c) sort((c).begin(),(c).end())
#define MIN(vec) *min_element(vec.begin(), vec.end())
#define MAX(vec) *max_element(vec.begin(), vec.end())
#define UNIQ(vec) vec.erase(unique(vec.begin(), vec.end()),vec.end()) //ソートの必要あり
#define IN(n,m)  (!(m.find(n) == m.end()))
#define ENUM(m) for (auto itr = m.begin(); itr != m.end(); ++itr)
#define dump_MAP(m) for(auto itr = m.begin(); itr != m.end(); ++itr) { cerr << itr->first << " --> "  << itr->second << endl; }
#define FINDL(vec,x) (lower_bound(vec.begin(),vec.end(),x) - vec.begin())
#define FINDU(vec,x) (upper_bound(vec.begin(),vec.end(),x) - vec.begin())
#define ROUND(N) setprecision(N)
#define ROUND_PRINT(N,val) cout << fixed;cout << setprecision(N) << val << endl
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define INARR(h,w,x,y) (0 <= y && y < h && 0 <= x && x < w)
#define EQ(a,b) (abs(a - b) < 1e-10)
using namespace std;
constexpr int dx[4] = {0,1,0,-1};
constexpr int dy[4] = {1,0,-1,0};
constexpr long double pi = M_PI;
constexpr double eps = 1e-10;
constexpr long mod = 1000000007;
constexpr short shINF = 32767;
constexpr long loINF = 2147483647;
constexpr long long llINF = 9223372036854775807;
typedef long long LL;
typedef vector<LL> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<LL,LL> pr;
typedef vector<bool> VB;
typedef vector<pr> VP;
struct Order {
	bool operator() (pr const& a,pr const& b) const {
		return a.first > b.first || ((a.first == b.first) && (a.second > b.second));
	}
};

template<typename T>
class Dijekstra {
private:
	struct ver {T di;int node,prev;};
	struct Order {
		bool operator() (ver const& a,ver const& b) const {
			return a.di > b.di || ((a.di == b.di) && (a.node > b.node));
		}
	};
	typedef priority_queue<ver,vector<ver>,Order> pq;
	vector<vector<pair<long,T>>> E;
	vector<long> trace;
	size_t V;
public:
	vector<T> dist;
	Dijekstra(size_t v) :
		V(v),E(v),dist(v,numeric_limits<T>::max()){}
	size_t size() { return V;}

	T & operator[] (int n) {return dist[n];}

	void add_edge_directed(long from,int to,T dis){
		E[from].push_back(pair<int,T>(to,dis));
	}

	void add_edge_undirected(int from,int to,T dis){
		E[from].push_back(pair<int,T>(to,dis));
		E[to].push_back(pair<int,T>(from,dis));
	}

	void dijekstra(int start){ // distを返す。
		pq que;
		dist[start] = 0;
		que.push(ver{0,start,start});

		while(! que.empty()){
			ver next = que.top();
			que.pop();
			int next_v = next.node;
			if (dist[next_v] < next.di) continue; //既により最適が見つかっている
			dist[next_v] = next.di;
			for(int i = 0;i < E[next_v].size();++i){
				pair<int,T> e = E[next_v][i];
				if (dist[e.first] == numeric_limits<T>::max()) que.push(ver{next.di+e.second,e.first,next_v});
			}
		}
	}

	vector<T> result(void){ return dist;}
};


int main(void){
	long long N,M,S,T,u,v,a,b;
	cin >> N >> M >> S >> T;
	Dijekstra<long long> yens(N);
	Dijekstra<long long> snookg(N);
	REP(i,M) {
		cin >> u >> v >> a >> b;
		yens.add_edge_undirected(u-1,v-1,a);
		snookg.add_edge_undirected(u-1,v-1,b);
	}
	yens.dijekstra(S-1);
	snookg.dijekstra(T-1);
	VI acc(N);
	acc[N-1] = yens[N-1] + snookg[N-1];
	REP(i,N-1) {
		acc[N-i-2] = yens[N-i-2]+snookg[N-i-2];
		acc[N-i-2] = min(acc[N-i-1],acc[N-i-2]);
	}
	long long ans = 1000000000000000;
	REP(i,N) cout << ans - (LL)acc[i] << endl;
}