//SPFA-标程
#include <bits/stdc++.h>
#define LL long long
#define INF 1e18
#define MAXN 100005
 
LL min(LL x, LL y) {if(x<y) return x; return y;}
 
int n, m, s, t, u, v;
LL a, b, f=1e15, ans[MAXN];
 
struct Graph
{
	int cnt, head[MAXN], vis[MAXN];
	LL dis[MAXN];
	
	struct Edge {int next, to; LL dis;} edge[MAXN*2];
	
	std::queue<int> q;
	
	void addedge(int from, int to, LL dis)
	{
		edge[++cnt].next=head[from];
		edge[cnt].to=to;
		edge[cnt].dis=dis;
		head[from]=cnt;
	}
	
	void spfa(int s)
	{
		for(int i=1; i<=n; i++) dis[i]=INF;
		memset(vis, 0, sizeof(vis));
		q.push(s); vis[s]=1; dis[s]=0;
		while(!q.empty())
		{
			int x=q.front(); q.pop();
			vis[x]=0;
			for(int i=head[x]; i; i=edge[i].next)
			{
				int to=edge[i].to;
				if(dis[to]>dis[x]+edge[i].dis)
				{
					dis[to]=dis[x]+edge[i].dis;
					if(!vis[to]) q.push(to), vis[to]=1;
				}
			}
		}
	}
} g1, g2;
 
int main()
{
	
	scanf("%d%d%d%d", &n, &m, &s, &t);
	for(int i=1; i<=m; i++)
	{
		scanf("%d%d%lld%lld", &u, &v, &a, &b);
		g1.addedge(u, v, a); g1.addedge(v, u, a);
		g2.addedge(v, u, b); g2.addedge(u, v, b);
	}
	g1.spfa(s);  g2.spfa(t);
	ans[n+1]=INF;
	for(int i=n; i>=1; i--) ans[i]=min(ans[i+1], g1.dis[i]+g2.dis[i]);
	for(int i=1; i<=n; i++) printf("%lld\n", f-ans[i]);
}
### Prompt

Generate a cpp solution to the following problem:

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
//SPFA-标程
#include <bits/stdc++.h>
#define LL long long
#define INF 1e18
#define MAXN 100005
 
LL min(LL x, LL y) {if(x<y) return x; return y;}
 
int n, m, s, t, u, v;
LL a, b, f=1e15, ans[MAXN];
 
struct Graph
{
	int cnt, head[MAXN], vis[MAXN];
	LL dis[MAXN];
	
	struct Edge {int next, to; LL dis;} edge[MAXN*2];
	
	std::queue<int> q;
	
	void addedge(int from, int to, LL dis)
	{
		edge[++cnt].next=head[from];
		edge[cnt].to=to;
		edge[cnt].dis=dis;
		head[from]=cnt;
	}
	
	void spfa(int s)
	{
		for(int i=1; i<=n; i++) dis[i]=INF;
		memset(vis, 0, sizeof(vis));
		q.push(s); vis[s]=1; dis[s]=0;
		while(!q.empty())
		{
			int x=q.front(); q.pop();
			vis[x]=0;
			for(int i=head[x]; i; i=edge[i].next)
			{
				int to=edge[i].to;
				if(dis[to]>dis[x]+edge[i].dis)
				{
					dis[to]=dis[x]+edge[i].dis;
					if(!vis[to]) q.push(to), vis[to]=1;
				}
			}
		}
	}
} g1, g2;
 
int main()
{
	
	scanf("%d%d%d%d", &n, &m, &s, &t);
	for(int i=1; i<=m; i++)
	{
		scanf("%d%d%lld%lld", &u, &v, &a, &b);
		g1.addedge(u, v, a); g1.addedge(v, u, a);
		g2.addedge(v, u, b); g2.addedge(u, v, b);
	}
	g1.spfa(s);  g2.spfa(t);
	ans[n+1]=INF;
	for(int i=n; i>=1; i--) ans[i]=min(ans[i+1], g1.dis[i]+g2.dis[i]);
	for(int i=1; i<=n; i++) printf("%lld\n", f-ans[i]);
}
```