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;
long long dis1[100001],dis2[100001],ans[100001];
vector<pair<long long,long long> > adj1[100001],adj2[100001];
int main(){
	int n,m,s,t;
	scanf("%d%d%d%d",&n,&m,&s,&t);
	while (m--){
		 long long x,y,a,b;
		 scanf("%lld%lld%lld%lld",&x,&y,&a,&b);
		 adj1[x].push_back(make_pair(y,a));adj1[y].push_back(make_pair(x,a));
		 adj2[x].push_back(make_pair(y,b));adj2[y].push_back(make_pair(x,b));
	}
	memset(dis1,0x7f,sizeof(dis1));
	memset(dis2,0x7f,sizeof(dis2));
	priority_queue<pair<long long,long long> > q;
	dis1[s]=0;
	q.push(make_pair(0,s));
	while (!q.empty()){
		long long x=q.top().second;q.pop();
		for (int i=0;i<adj1[x].size();i++){
			long long y=adj1[x][i].first,z=adj1[x][i].second+dis1[x];
			if (dis1[y]>z){
				dis1[y]=z;
				q.push(make_pair(-dis1[y],y));
			}
		}
	}
	dis2[t]=0;
	q.push(make_pair(0,t));
	while (!q.empty()){
		long long x=q.top().second;q.pop();
		for (int i=0;i<adj2[x].size();i++){
			long long y=adj2[x][i].first,z=adj2[x][i].second+dis2[x];
			if (dis2[y]>z){
				dis2[y]=z;
				q.push(make_pair(-dis2[y],y));
			}
		}
	}
	ans[n+1]=dis1[0];
	for (int i=n;i;i--) ans[i]=min(ans[i+1],dis1[i]+dis2[i]);
	for (int i=1;i<=n;i++) printf("%lld\n",((long long)(1e15+1e-8))-ans[i]);
	return 0;
}