Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

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

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

struct P
{
    int x,y,z;

    bool operator<(const P &a)const{
              return x>a.x;
    }

};

vector<P> v[101103];
//bitset<4001000> b;
int a,c,i,b,k,n,d,m;//dy[15]={0,1,0,-1,-1,1,-1,1},dx[15]={1,0,-1,0,1,1,-1,-1};//
long long o[100130];
long long l[101103];
int j[101130];
int e[110];
int dx[10]={1,0,-1,0,1,1,-1,-1},dy[10]={0,1,0,-1,1,-1,1,-1},dz[10]={0,0,0,0,1,-1};
long long x,y,mod=998244353;
long long z;
double pi=3.14159265;
P u[10600];
stack<int> s;

//set<int> s;
queue<int> q;

//2147483647
//'1'==49;
//'A'==65;
//'a'==97;
//unordered_
//map<int,int> p,p1;
//list<int> l;

string r;
//char r[133];
//deque<int> de;
bool as(P a,P b)
{
    if(o[a.x]!=o[b.x])
    return o[a.x]<o[b.x];
    return a.y<b.y;
}


int main()
{
    scanf("%d %d %d %d",&a,&b,&c,&d);
    for(int t=1;t<=b;t++)
    {
        scanf("%d %d %d %d",&n,&m,&i,&k);
        v[n].push_back({m,i,k});
        v[m].push_back({n,i,k});
    }
    memset(o,-1,sizeof(o));
    memset(l,-1,sizeof(l));
    q.push(c);
    j[c]=1;
    o[c]=0;
    for(;q.size();j[q.front()]=0,q.pop())
    for(int h=0;h<v[q.front()].size();h++)
    if(o[v[q.front()][h].x]<0||o[v[q.front()][h].x]>o[q.front()]+v[q.front()][h].y)
    {
        o[v[q.front()][h].x]=o[q.front()]+v[q.front()][h].y;
        if(!j[v[q.front()][h].x])
        {
            j[v[q.front()][h].x]=1;
            q.push(v[q.front()][h].x);
        }
    }


    q.push(d);
    j[d]=1;
    l[d]=0;
    for(;q.size();j[q.front()]=0,q.pop())
    for(int h=0;h<v[q.front()].size();h++)
    if(l[v[q.front()][h].x]<0||l[v[q.front()][h].x]>l[q.front()]+v[q.front()][h].z)
    {
        l[v[q.front()][h].x]=l[q.front()]+v[q.front()][h].z;
        if(!j[v[q.front()][h].x])
        {
            j[v[q.front()][h].x]=1;
            q.push(v[q.front()][h].x);
        }
    }
    for(int t=1;t<=a;t++)
        o[t]+=l[t];
    for(int t=a-1;t>0;t--)
        o[t]=min(o[t],o[t+1]);
    for(int t=1;t<=a;t++)
        printf("%lld\n",1000000000000000-o[t]);
}