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<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;

#define MAXN 100000
int N,M;
struct Edge{
    int v,w[2];
    Edge(){}
    Edge(int a,int w1,int w2){
        v=a,w[0]=w1,w[1]=w2;
    }
};
vector<Edge> G[MAXN+5];

bool vis[MAXN+5];
long long dist[2][MAXN+5];
void SPFA(int S,int opt){
    memset(vis,0,sizeof vis);
    memset(dist[opt],0x3f,sizeof dist[opt]);
    queue<int> Q;Q.push(S);
    dist[opt][S]=0;vis[S]=1;
    while(!Q.empty()){
        int u=Q.front();
        vis[u]=0;Q.pop();
        for(int i=0;i<int(G[u].size());i++){
            int v=G[u][i].v,w=G[u][i].w[opt];
            if(dist[opt][v]>dist[opt][u]+w){
                dist[opt][v]=dist[opt][u]+w;
                if(!vis[v]){
                    vis[v]=1;
                    Q.push(v);
                }
            }
        }
    }
}

long long Ans[MAXN+5];
int main(){
    int S,T;
    scanf("%d%d%d%d",&N,&M,&S,&T);
    for(int i=1;i<=M;i++){
        int u,v,w1,w2;
        scanf("%d%d%d%d",&u,&v,&w1,&w2);
        G[u].push_back(Edge(v,w1,w2));
        G[v].push_back(Edge(u,w1,w2));
    }
    SPFA(S,0);
    SPFA(T,1);
    Ans[N]=dist[0][N]+dist[1][N];
    for(int i=N-1;i>=1;i--)
        Ans[i]=min(Ans[i+1],dist[0][i]+dist[1][i]);
    for(int i=1;i<=N;i++)
        printf("%lld\n",1000000000000000ll-Ans[i]);
}
