Problem:
Score : 500 points

Problem Statement

There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes.
You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin.
The transaction takes D_i minutes for each gold coin you give.
You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq 100

0 \leq S \leq 10^9

1 \leq A_i \leq 50

1 \leq B_i,C_i,D_i \leq 10^9

1 \leq U_i < V_i \leq N

There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).

Each city t=2,...,N can be reached from City 1 with some number of railroads.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N

Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Sample Input 1

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5

Sample Output 1

2
14

The railway network in this input is shown in the figure below.

In this figure, each city is labeled as follows:

The first line: the ID number i of the city (i for City i)

The second line: C_i / D_i

Similarly, each railroad is labeled as follows:

The first line: the ID number i of the railroad (i for the i-th railroad in input)

The second line: A_i / B_i

You can travel from City 1 to City 2 in 2 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

You can travel from City 1 to City 3 in 14 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 6 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Sample Input 2

4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4

Sample Output 2

5
5
7

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 4 in 7 minutes, as follows:

At the exchange counter in City 1, exchange 2 gold coins for 6 silver coins in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Use the 4-th railroad to move from City 3 to City 4 in 1 minutes.

Sample Input 3

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1

Sample Output 3

1
9003
14606
16510
16576

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 6 in 16576 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 9000 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

At the exchange counter in City 3, exchange 8 gold coins for 8 silver coins in 5600 minutes.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Use the 3-rd railroad to move from City 2 to City 4 in 1 minute.

At the exchange counter in City 4, exchange 19 gold coins for 19 silver coins in 1900 minutes.

Use the 3-rd railroad to move from City 4 to City 2 in 1 minute.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

Use the 4-th railroad to move from City 3 to City 5 in 1 minute.

At the exchange counter in City 5, exchange 63 gold coins for 63 silver coins in 63 minutes.

Use the 4-th railroad to move from City 5 to City 3 in 1 minute.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 5-th railroad to move from City 1 to City 6 in 1 minute.

Sample Input 4

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7

Sample Output 4

1
3
5

The railway network in this input is shown in the figure below:

Sample Input 5

2 1 0
1 2 1 1
1 1000000000
1 1

Sample Output 5

1000000001

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 2 in 1000000001 minutes, as follows:

At the exchange counter in City 1, exchange 1 gold coin for 1 silver coin in 1000000000 minutes.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class Main implements Runnable
{
//    static int dx[]={-1,0,1,0};
//    static int dy[]={0,-1,0,1};
    static long mod=1000000007;
    
    public static void main(String[] args) {        
        new Thread(null, new Main(), "", 200 * 1024 * 1024).start();        
    }

    public void run() {
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);

        int n=sc.nextInt();
        int m=sc.nextInt();
        int s=sc.nextInt();
        ArrayList<Integer>[] l=new ArrayList[n];
        for (int i = 0; i < l.length; i++) {
			l[i]=new ArrayList<>();
		}
        int sf=10,bt=(1<<sf)-1;
        int e[][]=new int[m][2];
        int ex[][]=new int[n][2];
        for (int i = 0; i < m; i++) {
			int f=sc.nextInt()-1,t=sc.nextInt()-1;
			l[f].add((t<<sf)+i);
			l[t].add((f<<sf)+i);
			e[i][0]=sc.nextInt();
			e[i][1]=sc.nextInt();
		}
        for (int i = 0; i < ex.length; i++) {
			ex[i][0]=sc.nextInt();
			ex[i][1]=sc.nextInt();
		}
        
        long d[][]=new long[50][3000];
        for (int i = 0; i < d.length; i++) {
			fill(d[i], 1L<<50);
		}
        
        LinkedList<P> q=new LinkedList<>();
        d[0][s]=0;
        q.add(new P(0,s,0));

        while(!q.isEmpty()) {
        	P p=q.poll();
//        	db(p.x,p.c,p.s);
        	if(p.s>d[p.x][p.c])continue;
			if(p.c+ex[p.x][0]<3000 && p.s+ex[p.x][1] < d[p.x][p.c+ex[p.x][0]]) {
				d[p.x][ex[p.x][0]] = p.s + ex[p.x][1];
				q.add(new P(p.x, p.c + ex[p.x][0], p.s + ex[p.x][1]));
			}
			for (int i = 0,to,id; i < l[p.x].size(); i++) {
				id = l[p.x].get(i);
				to=id>>sf;
				id=id&bt;
				if(p.c < e[id][0])continue;
				if(p.s + e[id][1] < d[to][p.c-e[id][0]]) {
					d[to][p.c-e[id][0]] = p.s + e[id][1];
					q.add(new P(to, p.c - e[id][0], p.s + e[id][1]));
				}
			}
        }
        for (int i = 1; i < n; i++) {
        	sort(d[i]);
			out.println(d[i][0]);
		}
    
        out.println();
     	out.flush();
    }

    static void ret(String ans) {
    	System.out.println(ans);
    	System.exit(0);
    }

    static void ret(long ans) {
    	System.out.println(ans);
    	System.exit(0);
    }
    
    static boolean validpos(int x,int y,int r, int c){
        return x<r && 0<=x && y<c && 0<=y;
    }

    static void db(Object... os){
        System.err.println(Arrays.deepToString(os));
    }
}

class P {
	int x,c;long s;
	P(int  x, int c, long s) {
	  this.x=x;
	  this.c=c;
	  this.s=s;
	}
}


//class P implements Comparable<P>{
//    int i,x,r,f;
//    P(int i,int x,int r,int f) {
//    	this.i=i;
//    	this.x=x;
//    	this.r=r;
//    	this.f=f;
//    }
//
//    public int compareTo(P p){
//    	if(-this.r+p.r==0) {
//    		return -this.x+p.x;
//    	}
//        return -this.r+p.r;
//    }
//}

class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString()
    {
        while( st==null || !st.hasMoreTokens() )
			try {
				st = new StringTokenizer(x.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
        return st.nextToken();
    }
    public int nextInt()
    {
        return Integer.parseInt(nextString());
    }
    public int[] nextIntArray(int size) {
        int r[] = new int[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextInt(); 
        }
        return r;
    }
    public long[] nextLongArray(int size) {
        long r[] = new long[size];
        for (int i = 0; i < size; i++) {
            r[i] = this.nextLong(); 
        }
        return r;
    }
    public char[] getCharSet() {
        return this.nextString().toCharArray();
    }    
    public long nextLong()
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}