#include<bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const ll MOD=1000000007;
const ll INF=1000000010;
const ll LINF=4000000000000000010;
int dx[5]={0,1,0,-1,0};
int dy[5]={1,0,-1,0,0};
struct edge{int to;ll cost;};
vector<ll> dijkstra(int s,vector<edge> g[]){
    vector<ll> d(100010,LINF);
    priority_queue<P,vector<P>,greater<P>> q;
    d[s]=0;
    q.push(P(0,s));
    while(!q.empty()){
        P p=q.top();q.pop();
        int v=p.second;
        for(int i=0;i<g[v].size();i++){
            edge e=g[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                q.push(P(d[e.to],e.to));
            }
        }
    }
    return d;
}
vector<edge> G1[100010],G2[100010];
vector<ll> d1,d2;
int main(){
    int n,m,s,t;cin>>n>>m>>s>>t;s--;t--;
    while(m--){
        int u,v;cin>>u>>v;u--;v--;
        ll a,b;cin>>a>>b;
        G1[u].push_back({v,a});
        G2[u].push_back({v,b});
        G1[v].push_back({u,a});
        G2[v].push_back({u,b});
    }
    d1=dijkstra(s,G1);
    d2=dijkstra(t,G2);
    ll mi[100010];
    fill(mi,mi+n+2,LINF);
    for(int i=n-1;i>=0;i--){
        ll s=d1[i]+d2[i];
        mi[i]=min(mi[i+1],s);
    }
    for(int i=0;i<n;i++){
        cout<<1000000000000000-mi[i]<<endl;
    }
}

### Prompt

Develop a solution in Cpp 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<bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const ll MOD=1000000007;
const ll INF=1000000010;
const ll LINF=4000000000000000010;
int dx[5]={0,1,0,-1,0};
int dy[5]={1,0,-1,0,0};
struct edge{int to;ll cost;};
vector<ll> dijkstra(int s,vector<edge> g[]){
    vector<ll> d(100010,LINF);
    priority_queue<P,vector<P>,greater<P>> q;
    d[s]=0;
    q.push(P(0,s));
    while(!q.empty()){
        P p=q.top();q.pop();
        int v=p.second;
        for(int i=0;i<g[v].size();i++){
            edge e=g[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                q.push(P(d[e.to],e.to));
            }
        }
    }
    return d;
}
vector<edge> G1[100010],G2[100010];
vector<ll> d1,d2;
int main(){
    int n,m,s,t;cin>>n>>m>>s>>t;s--;t--;
    while(m--){
        int u,v;cin>>u>>v;u--;v--;
        ll a,b;cin>>a>>b;
        G1[u].push_back({v,a});
        G2[u].push_back({v,b});
        G1[v].push_back({u,a});
        G2[v].push_back({u,b});
    }
    d1=dijkstra(s,G1);
    d2=dijkstra(t,G2);
    ll mi[100010];
    fill(mi,mi+n+2,LINF);
    for(int i=n-1;i>=0;i--){
        ll s=d1[i]+d2[i];
        mi[i]=min(mi[i+1],s);
    }
    for(int i=0;i<n;i++){
        cout<<1000000000000000-mi[i]<<endl;
    }
}

```