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;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define range(a) (a).begin(),(a).end()
#define print(A,x,n) rep(i,0,n){cout<<(i ? " ":"")<<A[i]x;}cout<<endl;
#define pprint(A,y,m,n) rep(j,0,m){print(A[j],y,n);}
const long mod=1e9+7;
const int size=1e5;
const long inf=1e15;
int n,m;int s,t;
long d[2*size];
vector<pair<int,long>> g[2*size];
void dijkstra(){// O(NlogN+M)
	priority_queue<pair<long,int>> pq;
	rep(i,0,2*n){
		d[i] = inf;
		pq.push({-d[i] , i});
	}
	d[s] = 0; pq.push({0, s});
	d[t+n] = 0; pq.push({0, t+n});
	int v; long temp;
	while(!pq.empty()){
		v = pq.top().second;
		pq.pop();
        int u;
		for(pair<int,long> p:g[v]){
            u = p.first;
			temp = d[v] + p.second;
			if(d[u] > temp){
				d[u] = temp;
				pq.push({-d[u], u});
			}
		}
	}return;
}
int main(){
	cin>>n>>m; cin>>s>>t; s--; t--;
	int u,v; long a,b;
	rep(i,0,m){
		cin>>v>>u>>a>>b; v--; u--;
		g[v].push_back({u, a});
		g[u].push_back({v, a});
		g[v+n].push_back({u+n, b});
		g[u+n].push_back({v+n, b});
	}
	dijkstra();
	long ans[n+1]; ans[n] = inf;
	for(int i = n-1; i >= 0; i--){
		ans[i] = min(ans[i+1], d[i] + d[i+n]);
	}rep(i,0,n){
		cout<<inf - ans[i]<<endl;
	}
}