#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
#define MAXN 100006
#define INF 1000000000000000
using namespace std;
int n,m,s,t,u,v;
long long d[MAXN][2],ans[MAXN];
struct node
{
	long long v[2];
	int to;
};
vector<node> road[MAXN];
bool in[MAXN];
void SPFA(int x,int op)
{
	for(int i=0;i<MAXN;i++)
		d[i][op]=INF;
	d[x][op]=0;
	queue<int> q;
	q.push(x);
	memset(in,0,sizeof in);
	in[x]=1;
	while(!q.empty())
	{
		int now=q.front();
		q.pop();
		in[now]=0;
		for(int j=0;j<(int)road[now].size();j++)
		{
			int nex=road[now][j].to;
			if(d[nex][op]>d[now][op]+road[now][j].v[op])
			{
				d[nex][op]=d[now][op]+road[now][j].v[op];
				if(!in[nex])
				{
					in[nex]=1;
					q.push(nex);
				}
			}
		}
	}
}
int main()
{
	scanf("%d%d%d%d",&n,&m,&s,&t);
	node now;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%lld%lld",&u,&v,&now.v[0],&now.v[1]);
		now.to=v;
		road[u].push_back(now);
		now.to=u;
		road[v].push_back(now);
	}
	SPFA(s,0);
	SPFA(t,1);
	ans[n]=INF;
	for(int i=n-1;i>=0;i--)
		ans[i]=min(ans[i+1],d[i+1][0]+d[i+1][1]);
	for(int i=0;i<=n-1;i++)
		printf("%lld\n",INF-ans[i]);
}
### Prompt

Create a solution in cpp for 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
#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
#define MAXN 100006
#define INF 1000000000000000
using namespace std;
int n,m,s,t,u,v;
long long d[MAXN][2],ans[MAXN];
struct node
{
	long long v[2];
	int to;
};
vector<node> road[MAXN];
bool in[MAXN];
void SPFA(int x,int op)
{
	for(int i=0;i<MAXN;i++)
		d[i][op]=INF;
	d[x][op]=0;
	queue<int> q;
	q.push(x);
	memset(in,0,sizeof in);
	in[x]=1;
	while(!q.empty())
	{
		int now=q.front();
		q.pop();
		in[now]=0;
		for(int j=0;j<(int)road[now].size();j++)
		{
			int nex=road[now][j].to;
			if(d[nex][op]>d[now][op]+road[now][j].v[op])
			{
				d[nex][op]=d[now][op]+road[now][j].v[op];
				if(!in[nex])
				{
					in[nex]=1;
					q.push(nex);
				}
			}
		}
	}
}
int main()
{
	scanf("%d%d%d%d",&n,&m,&s,&t);
	node now;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%lld%lld",&u,&v,&now.v[0],&now.v[1]);
		now.to=v;
		road[u].push_back(now);
		now.to=u;
		road[v].push_back(now);
	}
	SPFA(s,0);
	SPFA(t,1);
	ans[n]=INF;
	for(int i=n-1;i>=0;i--)
		ans[i]=min(ans[i+1],d[i+1][0]+d[i+1][1]);
	for(int i=0;i<=n-1;i++)
		printf("%lld\n",INF-ans[i]);
}
```