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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
#include<assert.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")
#define charsize 100005 //10^5+5

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a= c;}
// void fill(ll n, ll a[n], ll value){ rep(i,0,n) a[i] = value; }
void cin(ll *n){ scanf("%lld",&(*n)); }
void cin2(ll *a, ll *b){ scanf("%lld%lld",&(*a),&(*b)); }
void cin3(ll *a, ll *b, ll *c){ scanf("%lld%lld%lld",&(*a),&(*b),&(*c)); }
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[n]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[n]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll ABS(ll a){return a>=0?a:(-a);}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod_MOD1(ll n){n+= n<0?((-n)/MOD1+1)*MOD1:0; return n%=MOD1;}
ll mod_p(ll n ,ll p){n+= n<0?((-n)/p+1)*p:0; return n%=p;}
ll change_into_num(char s[] , ll len, ll p){ return !p ? 0 : POW(10,p-1)*(s[len-p]-'0') + change_into_num(s,len,p-1); }

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}

typedef struct{ ll a , b;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->a - ((fr*)q)->a;}
int cmp2( const void *p, const void *q ) { return ((fr*)q) ->a - ((fr*)p)->a;}
void strsortup(fr*a,int n){qsort(a,n,sizeof(fr),cmp1);}
void strsortdown(fr*a,int n){qsort(a,n,sizeof(fr),cmp2);}

/*------------------------------Dijkstra start-------------------------------------------------------------*/
//http://www.deqnotes.net/acmicpc/dijkstra/
int edge_cnt[114123];
ll* graph[114123];
ll ans[114123];
bool is_conf[114123];
// freeeeeeeeeeeeeeeeeeeeeeeeeeee!!!!!!!!!!!!

void pushGraph(ll v1, ll v2, ll weight , ll is_dir , ll snuke){
    graph[v1] = realloc( graph[v1] , sizeof(ll)*(edge_cnt[v1]*3+3) );    
    graph[v1][edge_cnt[v1]*3] = v2;
    graph[v1][edge_cnt[v1]*3+1] = weight;
    graph[v1][edge_cnt[v1]*3+2] = snuke;
    // ednum[v1] = realloc( ednum[v1] , sizeof(ll)*(edge_cnt[v1]+1) );
    // ednum[v1][edge_cnt[v1]] = num;
    edge_cnt[v1]++;

    if(!is_dir){
        graph[v2] = realloc( graph[v2] , sizeof(ll)*(edge_cnt[v2]*3+3) );
        graph[v2][edge_cnt[v2]*3] = v1;
        graph[v2][edge_cnt[v2]*3+1] = weight;
        graph[v2][edge_cnt[v2]*3+2] = snuke;
        // ednum[v2] = realloc( ednum[v2] , sizeof(ll)*(edge_cnt[v2]+1) );
        // ednum[v2][edge_cnt[v2]] = num;
        edge_cnt[v2]++;
    }
}
/*-----------priorityqueue(strurct)---------------*/
// kind = max?1:0
ll HeapValue(ll a){return a?-INF:INF;}
ll Heapcmp(ll a, ll b, ll kind){return kind ? (a<b?1:0) : (a>b?1:0);}
typedef struct{ int from; int to; ll cost; }graph_heap;
graph_heap heap[2*MAX_N];
void str_swap(graph_heap *a,graph_heap *b){ graph_heap c;c=*b;*b=*a;*a=c; }

void graph_pushHeap(ll val, ll len , ll kind){
    ll i=len; heap[i].cost=val;
    while(i!=0){
        if( Heapcmp( heap[i].cost, heap[(i-1)/2].cost , ABS(kind-1) )  ) str_swap(&heap[i],&heap[(i-1)/2]);
        i--;i/=2;
    }
}
void graph_popHeap(ll len, ll kind){
    ll i=0; heap[0].cost=HeapValue(kind);
    str_swap(&heap[0],&heap[len-1]);
    while(len>i*2+1){
        if(i*2+2<len && Heapcmp(heap[i].cost,heap[i*2+2].cost,kind) && Heapcmp(heap[i*2+1].cost,heap[i*2+2].cost,kind) ){
            str_swap(&heap[i],&heap[i*2+2]);
            i*=2;i+=2;
        }else{
            if( Heapcmp(heap[i].cost,heap[i*2+1].cost,kind) ){
                str_swap(&heap[i],&heap[i*2+1]);
                i*=2;i++;
            }else break;
        }
    }
}
/*---------------------------------------------*/

void dijkstra(ll start, ll snuke){
    rep(i,0,114123){
        ans[i] = INF;
        is_conf[i] = false;
    }
    ans[start] = 0;
    is_conf[start]  = true;
    int size=0;
    rep(i,0,edge_cnt[start]){
        if( snuke != graph[start][i*3+2] ) continue;
        heap[size].from = start;
        heap[size].to = graph[start][i*3];
        ans[ heap[size].to ] = graph[start][i*3+1];
        graph_pushHeap( graph[start][i*3+1] , size , 0 );
        size++;        
    }
    while(size){
        if( is_conf[ heap[0].to ] ){
            graph_popHeap(size,0);
            size--;
            continue;
        }
        // printf("AA: %d %d\n",heap[0].from, heap[0].to );
        is_conf[ heap[0].to ] = true;
        ans[ heap[0].to ] = heap[0].cost;
        // rep(i,0,4) printf("%lld ",ans[i]);
        // PN;
        rep(i,0,edge_cnt[ heap[0].to ]){
            if( snuke != graph[ heap[0].to ][i*3+2] ) continue;
            if( ans[ graph[ heap[0].to ][i*3] ] > ans[ heap[0].to ] + graph[ heap[0].to ][i*3+1] ){
                heap[size].from = heap[0].to;
                heap[size].to = graph[ heap[0].to ][i*3];
                graph_pushHeap( ans[ heap[0].to ] + graph[ heap[0].to ][i*3+1] , size , 0 );
                size++;
            }
        }
        graph_popHeap(size,0);
        size--;
    }
}
/*---------------------------Dijkstra end--------------------------------------------------------------------*/


int main(void){
    // char ;
    ll n,m,s,t;
    // ll ans=0;
    cin(&n);
    cin(&m);
    cin2(&s,&t);
    // cin3(&,&,&);
    ll u,v,a,b;
    rep(i,0,m){
        cin2(&u,&v);
        cin2(&a,&b);
        pushGraph(u-1,v-1,a,0,0);
        pushGraph(u-1,v-1,b,0,1);
    }
    // PN;
    // rep(i,0,n){
    //     rep(j,0,edge_cnt[i]){
    //         printf("%lld %lld %lld %lld\n",i,graph[i][j*3],graph[i][j*3+1],graph[i][j*3+2] );
    //     }
    // }
    // PN;
    ll total_cost[n],answer[n],low=INF;
    rep(i,0,n) total_cost[i]=0;
    dijkstra(s-1,0);
    // rep(i,0,n) printf("%lld ",ans[i] );
    // PN;
    rep(i,0,n) total_cost[i]+=ans[i];
    dijkstra(t-1,1);
    // rep(i,0,n) printf("%lld ",ans[i] );
    // PN;
    rep(i,0,n) total_cost[i]+=ans[i];
    // rep(i,0,n) printf("%lld ",total_cost[i] );
    // PN;

    rrep(i,n-1,0){
        low = min2(low,total_cost[i]);
        answer[i] = low;
    }

    ll kenko = POW(10,15);
    rep(i,0,n){
        printf("%lld\n",kenko-answer[i]);
    }

    
    // printf("%lld\n");
    return 0;
}