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.

Examples

Input

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


Output

2
14


Input

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


Output

5
5
7


Input

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


Output

1
9003
14606
16510
16576


Input

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


Output

1
3
5


Input

2 1 0
1 2 1 1
1 1000000000
1 1


Output

1000000001
import java.io.*;
import java.util.*;

public class Main {
    public static final long mod = (long)1e9+7;
    public static final long INF = Long.MAX_VALUE/2;
    public static final int inf = Integer.MAX_VALUE/2;

    static void solve(InputReader in, PrintWriter out){
        int n = in.ni(), m = in.ni(), s = Math.min(in.ni(), 50*(n-1));
        Dijkstra dk = new Dijkstra(n*(50*(n-1)+1));  // Max{A_i} <= 50
        for (int i = 0; i < m; i++) {
            int u = in.ni()-1, v = in.ni()-1, a = in.ni();
            long b = in.nl();
            for (int j = 0; j <= 50*(n-1); j++) {
                if(j-a>=0){
                    dk.addEdge(j*n+u, (j-a)*n+v, b);
                    dk.addEdge(j*n+v, (j-a)*n+u, b);
                }
            }
        }
        for (int i = 0; i < n; i++) {
            int c = in.ni(), d = in.ni();
            for (int j = 0; j < 50*(n-1); j++) {
                if(j+c<=50*(n-1)) dk.addEdge(j*n+i, (j+c)*n+i, d);
                else dk.addEdge(j*n+i, 50*(n-1)*n+i, d);
            }
        }
        dk.dijkstra(s*n);
        StringBuilder sb = new StringBuilder();
        for (int t = 1; t <= n-1; t++) {
            long min = INF;
            for (int i = 0; i <= 50*(n-1); i++) {
                min = Math.min(min, dk.dist[i*n+t]);
            }
            sb.append(min).append("\n");
        }
        out.print(sb);
    }
    public static class Dijkstra {
        int n; long[] dist;
        ArrayList<Pair>[] list;
        final long inf = Long.MAX_VALUE/3;
        public Dijkstra(int n) {
            this.n = n;
            this.dist = new long[n];
            list = new ArrayList[n];
            for (int i = 0; i < n; i++) {
                list[i] = new ArrayList<>();
                dist[i] = inf;
            }
        }
        public void addEdge(int from, int to, long cost) {
            list[from].add(new Pair(to, cost));
        }
        public void dijkstra(int s) {
            PriorityQueue<Pair> Q = new PriorityQueue<>();
            Q.add(new Pair(s, 0));
            boolean[] used = new boolean[n];
            while (!Q.isEmpty()) {
                Pair p = Q.poll();
                if (used[p.to]) continue;
                used[p.to] = true;
                dist[p.to] = p.cost;
                for (Pair edge : list[p.to]) {
                    Q.add(new Pair(edge.to, p.cost + edge.cost));
                }
            }
        }
        class Pair implements Comparable<Pair> {
            int to; long cost;
            Pair(int to, long cost) {
                this.to = to;
                this.cost = cost;
            }
            public int compareTo(Pair p) {
                return Long.compare(cost, p.cost);
            }
        }
    }

    public static void main(String[] args){
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        solve(in, out);
        out.close();
    }
    public static class InputReader{
        private BufferedReader br;
        private StringTokenizer st;
        public InputReader(InputStream is){
            br = new BufferedReader(new InputStreamReader(is));
            st = null;
        }
        public String ns(){
            if(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (Exception e){
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }
        public int ni(){
            return Integer.parseInt(ns());
        }
        public long nl(){
            return Long.parseLong(ns());
        }
        public double nd(){
            return Double.parseDouble(ns());
        }
        public char nc(){
            return ns().toCharArray()[0];
        }
        public int[] ni(int n){
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }
        public long[] nl(int n){
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }
        public double[] nd(int n){
            double[] a = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }
    }
}