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;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vl;

#define REP(i,a,b) for(ll i=a;i<b;i++)
#define rep(i,n) REP(i,0,n)

template< typename Monoid >
struct SegmentTree {
  using F = function< Monoid(Monoid, Monoid) >;
 
  ll sz;
  vector< Monoid > seg;
 
  const F f;
  const Monoid M1;
 
  SegmentTree(ll n, const F f, const Monoid &M1) : f(f), M1(M1) {
    sz = 1;
    while(sz < n) sz <<= 1;
    seg.assign(2 * sz, M1);
  }
 
  void set(ll k, const Monoid &x) {
    seg[k + sz] = x;
  }
 
  void build() {
    for(ll k = sz - 1; k > 0; k--) {
      seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
    }
  }
 
  void update(ll k, const Monoid &x) {
    k += sz;
    seg[k] = x;
    while(k >>= 1) {
      seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
    }
  }
 
  Monoid query(ll a, ll b) {
    Monoid L = M1, R = M1;
    for(a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
      if(a & 1) L = f(L, seg[a++]);
      if(b & 1) R = f(seg[--b], R);
    }
    return f(L, R);
  }
 
  Monoid operator[](const ll &k) const {
    return seg[k + sz];
  }
};

const ll MAX_V = 100000;

struct edge{ll from; ll to; ll cost;};
vector<edge> G1[MAX_V];
ll d1[MAX_V];

void add_edge1(ll from, ll to, ll cost) {
    edge e = {from, to, cost};
    G1[from].push_back(e);
}

void dijkstra1(ll s) {
    struct q_v{
        ll id;
        ll d;
        bool operator< (const q_v &t) const {
            return t.d < d;
        }
    };
    bool end[MAX_V] = {};
    for (ll i = 0; i < MAX_V; i++) d1[i] = (ll) 1e18;
    d1[s] = 0;
    priority_queue<q_v> q;
    q.push(q_v{s, 0});
    while (!q.empty()) {
        ll v = q.top().id;
        q.pop();
        if (end[v]) continue;
        end[v] = true;
        for (edge e : G1[v]) {
            if (!end[e.to]) {
                d1[e.to] = min(d1[e.to], d1[v] + e.cost);
                q.push(q_v{e.to, d1[e.to]});
            }
        }
    }
}

vector<edge> G2[MAX_V];
ll d2[MAX_V];

void add_edge2(ll from, ll to, ll cost) {
    edge e = {from, to, cost};
    G2[from].push_back(e);
}

void dijkstra2(ll s) {
    struct q_v{
        ll id;
        ll d;
        bool operator< (const q_v &t) const {
            return t.d < d;
        }
    };
    bool end[MAX_V] = {};
    for (ll i = 0; i < MAX_V; i++) d2[i] = (ll) 1e18;
    d2[s] = 0;
    priority_queue<q_v> q;
    q.push(q_v{s, 0});
    while (!q.empty()) {
        ll v = q.top().id;
        q.pop();
        if (end[v]) continue;
        end[v] = true;
        for (edge e : G2[v]) {
            if (!end[e.to]) {
                d2[e.to] = min(d2[e.to], d2[v] + e.cost);
                q.push(q_v{e.to, d2[e.to]});
            }
        }
    }
}

ll n, m, s, t;
string S;
ll ans = (ll)1e18;

ll u[101010], v[101010];
ll a[101010], b[101010];

void solve() {
    cin >> n >> m >> s >> t;
    rep(i,m) {
        cin >> u[i] >> v[i] >> a[i] >> b[i];
        u[i]--; v[i]--;
    }
    rep(i,m) {
        add_edge1(u[i], v[i], a[i]);
        add_edge1(v[i], u[i], a[i]);
    }
    dijkstra1(s-1);
    rep(i,m) {
        add_edge2(u[i], v[i], b[i]);
        add_edge2(v[i], u[i], b[i]);
    }
    dijkstra2(t-1);
    SegmentTree< ll > seg(n, [](ll a, ll b) { return min(a, b); }, (ll)1e18);
    rep(i,n) {
        seg.update(i, d1[i] + d2[i]);
    }
    rep(i,n) {
        cout << (ll)1e15 - seg.query(i,n) << endl;
    }
}

int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);
    solve();
    return 0;
}