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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD=1e9+7;
const int INF=1e9;

typedef pair<ll,int> pli;

int main(){
    int n,m,s,t;
    cin>>n>>m>>s>>t;
    s--; t--;
    vector<pli> a[n],b[n];
    for(int i=0;i<m;i++){
        int u,v;
        ll aa,bb;
        cin>>u>>v>>aa>>bb;
        u--; v--;
        a[u].push_back({aa,v});
        a[v].push_back({aa,u});
        b[u].push_back({bb,v});
        b[v].push_back({bb,u});
    }
    ll d1[n],d2[n];
    for(int i=0;i<n;i++){
        d1[i]=d2[i]=1e18;
    }
    priority_queue<pli,vector<pli>,greater<pli>> pq;
    pq.push({0LL,s});
    d1[s]=0LL;
    while(pq.size()){
        auto v=pq.top();
        pq.pop();
        if(d1[v.second]<v.first)continue;
        for(auto u:a[v.second]){
            if(d1[u.second]>v.first+u.first){
                d1[u.second]=v.first+u.first;
                pq.push({d1[u.second],u.second});
            }
        }
    }
    pq.push({0LL,t});
    d2[t]=0LL;
    while(pq.size()){
        auto v=pq.top();
        pq.pop();
        if(d2[v.second]<v.first)continue;
        for(auto u:b[v.second]){
            if(d2[u.second]>v.first+u.first){
                d2[u.second]=v.first+u.first;
                pq.push({d2[u.second],u.second});
            }
        }
    }
    ll ans[n];
    ans[n-1]=d1[n-1]+d2[n-1];
    for(int i=n-2;i>=0;i--){
        ans[i]=min(ans[i+1],d1[i]+d2[i]);
    }
    for(int i=0;i<n;i++){
        cout<<1000000000000000-ans[i]<<endl;
    }
}
