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.

Examples

Input

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1


Output

999999999999998
999999999999989
999999999999979
999999999999897


Input

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


Output

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
#include <bits/stdc++.h>

using namespace std;

const int MAXN = 1e5 + 5;
typedef long long ll;
const ll MONEY = 1000000000000000LL;

struct edge{
	int u, a, b;
};

struct node{
	int u;
	ll w;
	bool operator<(const node &rhs) const{
		return w > rhs.w;
	}
};

vector<edge> adj[MAXN];
ll yen[MAXN], snuuk[MAXN], suf[MAXN];
int N, M, s, t;

void YEN(){
	for(int i=1; i<=N; ++i) yen[i] = -1;
	yen[s] = 0;
	node cur = (node){s, 0};
	priority_queue<node> Q;
	Q.push(cur);
	while(!Q.empty()){
		cur = Q.top(); Q.pop();
		int u = cur.u;
		for(edge it : adj[u]){
			int v = it.u, a = it.a;
			if(yen[u] + a < yen[v] || yen[v] == -1){
				yen[v] = yen[u] + a;
				Q.push((node){v, yen[v]});
			}
		}
	}
}

void SNUUK(){
	for(int i=1; i<=N; ++i) snuuk[i] = -1;
	snuuk[t] = 0;
	node cur = (node){t, 0};
	priority_queue<node> Q;
	Q.push(cur);
	while(!Q.empty()){
		cur = Q.top(); Q.pop();
		int u = cur.u;
		for(edge it : adj[u]){
			int v = it.u, b = it.b;
			if(snuuk[u] + b < snuuk[v] || snuuk[v] == -1){
				snuuk[v] = snuuk[u] + b;
				Q.push((node){v, snuuk[v]});
			}
		}
	}
}

void solve(){
	ll mx = -MONEY;
	for(int i=N; i>=1; --i) suf[i] = max(suf[i + 1], MONEY -  yen[i] - snuuk[i]);
	for(int i=1; i<=N; ++i) printf("%lld\n", suf[i]);
}

int main(){
	scanf("%d %d %d %d", &N, &M, &s, &t);
	while(M--){
		int u, v, a, b;
		scanf("%d %d %d %d", &u, &v, &a, &b);
		adj[u].push_back((edge){v, a, b});
		adj[v].push_back((edge){u, a, b});
	}
	YEN();
	SNUUK();
	solve();
	return 0;
}