#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
#define i64 long long int
#define ran 101111
int n,m,s,t;
vector<pair<int,int> > e[ran], e2[ran];
i64 d[ran], d2[ran];
priority_queue<pair<i64,int> > q;
i64 res[ran];
void dij(int x,i64 d[ran],vector<pair<int,int> > e[ran]){
	for(int i=1; i<=n; i++)d[i] = -1;
	d[x] = 0;
	q.push(make_pair(0LL,x));
	while(!q.empty()){
		i64 dis = -q.top().first;
		x = q.top().second;
		q.pop();
		if(dis != d[x])continue;
		for(auto it : e[x]){
			int y = it.first;
			i64 dis2 = dis + it.second;
			if(d[y] == -1 || dis2 < d[y]){
				d[y] = dis2;
				q.push(make_pair(-dis2, y));
			}
		}
	}
}
int main(){
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=0; i<m; i++){
		int x,y,z,w;
		scanf("%d%d%d%d",&x,&y,&z,&w);
		e[x].push_back(make_pair(y,z));
		e[y].push_back(make_pair(x,z));
		e2[x].push_back(make_pair(y,w));
		e2[y].push_back(make_pair(x,w));
	}
	dij(s,d,e);
	dij(t,d2,e2);
	for(int i=n; i>=1; i--){
		res[i] = d[i] + d2[i];
		if(i<n && res[i] > res[i+1])
			res[i] = res[i+1];
	}
	for(int i=1; i<=n; i++)
		printf("%lld\n",((i64)1e15) - res[i]);
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include<cstdio>
#include<vector>
#include<queue>
using namespace std;
#define i64 long long int
#define ran 101111
int n,m,s,t;
vector<pair<int,int> > e[ran], e2[ran];
i64 d[ran], d2[ran];
priority_queue<pair<i64,int> > q;
i64 res[ran];
void dij(int x,i64 d[ran],vector<pair<int,int> > e[ran]){
	for(int i=1; i<=n; i++)d[i] = -1;
	d[x] = 0;
	q.push(make_pair(0LL,x));
	while(!q.empty()){
		i64 dis = -q.top().first;
		x = q.top().second;
		q.pop();
		if(dis != d[x])continue;
		for(auto it : e[x]){
			int y = it.first;
			i64 dis2 = dis + it.second;
			if(d[y] == -1 || dis2 < d[y]){
				d[y] = dis2;
				q.push(make_pair(-dis2, y));
			}
		}
	}
}
int main(){
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=0; i<m; i++){
		int x,y,z,w;
		scanf("%d%d%d%d",&x,&y,&z,&w);
		e[x].push_back(make_pair(y,z));
		e[y].push_back(make_pair(x,z));
		e2[x].push_back(make_pair(y,w));
		e2[y].push_back(make_pair(x,w));
	}
	dij(s,d,e);
	dij(t,d2,e2);
	for(int i=n; i>=1; i--){
		res[i] = d[i] + d2[i];
		if(i<n && res[i] > res[i+1])
			res[i] = res[i+1];
	}
	for(int i=1; i<=n; i++)
		printf("%lld\n",((i64)1e15) - res[i]);
	return 0;
}

```