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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define mod 1000000007

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

ll binpow(ll x,ll y)/* (x^y)%p in O(log y) */{ll res=1;while (y > 0){if(y&1)res=(res*x);y = y>>1;x=(x*x);}return res;}
ll binpowmod(ll x,ll y,ll p)/* (x^y)%p in O(log y) */{ull res=1;x=x%p;while (y > 0){if(y&1)res=(res*x)%p;y = y>>1;x=(x*x)%p;}return res;}
ll mod_inverse(ll n,ll p)/* Returns n^(-1) mod p */{return binpowmod(n,p-2,p);}
ll gcd(ll x,ll y)
{
    if(y==0)
        return x;
    return gcd(y,x%y);
}
bool comp_pairs_by_s(pair<ll,ll> &x ,pair<ll,ll> &y)
{
    return x.second<y.second;
}
bool isPowerOfTwo (ll x)  
{  
    /* First x in the below expression is for the case when x is 0 */
    return x && (!(x&(x-1)));  
}

class cmp      //comparator for priority_queue 
{               //declaration: priority_queue<int,vector<int>,cmp>
public:         
    bool operator()(pair<int,int> A,pair<int,int> B)
    {
        if(abs(A.first-A.second)==abs(B.first-B.second))
            return A.first>B.first;
        return abs(A.first-A.second)<abs(B.first-B.second);
    }
};
// int prime[10000007]={0};
// void sieve(void)
// {
//  int i,j;
//  for(i=0;i<10000007;i++)
//         prime[i]=1;
//  prime[0]=0,prime[1]=0;
//  for(i=2;i<=sqrt(10000007);i++){
//      if(prime[i]==i){
//          for(j=i*i;j<10000007;j+=i){
//             if(!prime[j])
//                 prime[j]=0;
//          }
//      }
//  }
    
// }
void swap(ll &x,ll &y){
    int temp=x;
    x=y;
    y=temp;
}
  
unsigned int onesComplement(unsigned int n) 
{ 
   // Find number of bits in the given integer 
   int number_of_bits = floor(log2(n))+1; 
  
   // XOR the given integer with poe(2,  
   // number_of_bits-1 and print the result  
   return ((1 << number_of_bits) - 1) ^ n; 
}
map<pair<ll,ll>,pair<ll,ll>> c;
vector<int> g[100001];
vector<bool> vis(100001,0);
vector<ll> yen(100001,INT_MAX);
vector<ll> snuke(100001,INT_MAX);
void solve()
{   
    int n,m,s,t;
    cin>>n>>m>>s>>t;
    for(int i=0;i<m;++i){
        ll u,v,a,b;
        cin>>u>>v>>a>>b;
        g[u].push_back(v); g[v].push_back(u);
        c[{u,v}]={a,b}; c[{v,u}]={a,b};
    }
    yen[s]=0;
    priority_queue<pair<ll,ll>> q;
    q.push({0,s});
    while(!q.empty())
    {
        pair<ll,ll> v=q.top(); q.pop();
        if(vis[v.second]) continue;
        vis[v.second]=1;
        yen[v.second]=v.first;
        for(int to:g[v.second])
        {
            if(yen[to]>yen[v.second]+c[{v.second,to}].first){
                yen[to]=min(yen[to],yen[v.second]+c[{v.second,to}].first);
                q.push({yen[to],to});
            }
        }
    }
    snuke[t]=0;
    for(int i=0;i<=n;++i) vis[i]=0;

    q.push({0,t});
    while(!q.empty())
    {
        pair<ll,ll> v=q.top(); q.pop();
         if(vis[v.second]) continue;
        vis[v.second]=1;
        snuke[v.second]=v.first;
        for(int to:g[v.second])
        {
            if(snuke[to]>snuke[v.second]+c[{v.second,to}].second){
                snuke[to]=min(snuke[to],snuke[v.second]+c[{v.second,to}].second);
                q.push({snuke[to],to});
            }
        }
    }
    // for(int i=1;i<=n;++i){
    //     cout<<yen[i]<<' '<<snuke[i]<<'\n';
    // }
    vector<pair<ll,ll>> tc(n);
    for(int i=0;i<n;++i){
        tc[i].first=yen[i+1]+snuke[i+1];
        tc[i].second=i+1;
    }
    sort(tc.begin(),tc.end());
    int j=0;
    debug(tc);
    ll total=1000000000000000;
    for(int i=0;i<n;++i){
        if(tc[i].second<=j) continue;
        while(j<tc[i].second){
            cout<<(total-tc[i].first)<<'\n'; ++j;
        }
    }
}   
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); 
    solve();
}