#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
#define LL long long
#define maxn 100005
using namespace std;
struct node { int v; LL l[2]; };
int n,m,s,t;
LL dist[2][maxn],ans[maxn];
vector<node> G[maxn];
queue<int> q;
bool inq[maxn];
void spfa(int s,int d)
{
	memset(dist[d],0x3f,sizeof(dist[d]));
	memset(inq,0,sizeof(inq));
	dist[d][s]=0;inq[s]=true;
	while(!q.empty()) q.pop();
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();q.pop();
		inq[u]=false;
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].v,di=G[u][i].l[d];
			if(dist[d][v]>dist[d][u]+di)
			{
				dist[d][v]=dist[d][u]+di;
				if(!inq[v]) { inq[v]=true; q.push(v); }
			}
		}
	}
}
int main()
{
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;i++)
	{
		int u;
		node p;
		scanf("%d%d%lld%lld",&u,&p.v,&p.l[0],&p.l[1]);
		G[u].push_back(p);
		swap(u,p.v);
		G[u].push_back(p);
	}
	spfa(s,0);spfa(t,1);
	ans[n]=dist[0][n]+dist[1][n];
	for(int i=n-1;i>0;i--) ans[i]=min(dist[0][i]+dist[1][i],ans[i+1]);
	for(int i=1;i<=n;i++) printf("%lld\n",(long long)(1e15)-ans[i]);
}
### Prompt

Please provide a cpp coded solution 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<queue>
#include<vector>
#include<cstring>
#include<algorithm>
#define LL long long
#define maxn 100005
using namespace std;
struct node { int v; LL l[2]; };
int n,m,s,t;
LL dist[2][maxn],ans[maxn];
vector<node> G[maxn];
queue<int> q;
bool inq[maxn];
void spfa(int s,int d)
{
	memset(dist[d],0x3f,sizeof(dist[d]));
	memset(inq,0,sizeof(inq));
	dist[d][s]=0;inq[s]=true;
	while(!q.empty()) q.pop();
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();q.pop();
		inq[u]=false;
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].v,di=G[u][i].l[d];
			if(dist[d][v]>dist[d][u]+di)
			{
				dist[d][v]=dist[d][u]+di;
				if(!inq[v]) { inq[v]=true; q.push(v); }
			}
		}
	}
}
int main()
{
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;i++)
	{
		int u;
		node p;
		scanf("%d%d%lld%lld",&u,&p.v,&p.l[0],&p.l[1]);
		G[u].push_back(p);
		swap(u,p.v);
		G[u].push_back(p);
	}
	spfa(s,0);spfa(t,1);
	ans[n]=dist[0][n]+dist[1][n];
	for(int i=n-1;i>0;i--) ans[i]=min(dist[0][i]+dist[1][i],ans[i+1]);
	for(int i=1;i<=n;i++) printf("%lld\n",(long long)(1e15)-ans[i]);
}
```