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.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;

import java.util.*;

public class Main {
    long MOD = (long) 1e9 + 7;

    InputStream in;
    OutputStream out;

    public Main() {
        this(System.in, System.out);
    }

    public Main(InputStream in, OutputStream out) {
        this.in = in;
        this.out = out;
    }

    int n, m, s;
    Edge[] g;
    int[] C;
    long[] D;

    void run() {
        Scanner sc = new Scanner(this.in);
        PrintWriter out = new PrintWriter(this.out);

        // write your code
        n = sc.nextInt();
        m = sc.nextInt();
        s = sc.nextInt();
        g = new Edge[n];
        for (int i = 0; i < n; i++) {
            g[i] = new Edge();
        }
        for (int i = 0; i < m; i++) {
            int u = sc.nextInt() - 1, v = sc.nextInt() - 1, a = sc.nextInt(), b = sc.nextInt();
            g[u].add(new Node(v, a, b));
            g[v].add(new Node(u, a, b));
        }

        C = new int[n];
        D = new long[n];
        for (int i = 0; i < n; i++) {
            C[i] = sc.nextInt();
            D[i] = sc.nextInt();
        }

        long[][] d = dijkstra();

        for (int i = 1; i < n; i++) {
            long min = INF;
            for (long v : d[i]) {
                min = Math.min(min, v);
            }
            out.println(min);
        }

        out.flush();
    }

    long INF = 1L << 60;
    int MAX = 2500;

    long[][] dijkstra() {
        long[][] d = new long[n][MAX + 1];
        for (long[] a : d)
            Arrays.fill(a, INF);

        PriorityQueue<Value> q = new PriorityQueue<>();

        int s = Math.min(MAX, this.s);
        for (int i = 0; s + i * C[0] <= MAX; i++) {
            d[0][s + i * C[0]] = i * D[0];
            q.add(new Value(0, i * D[0], s + i * C[0]));
        }
        HashSet<Integer> visited = new HashSet<>();

        while (!q.isEmpty()) {
            Value cur = q.remove();
            if (cur.v != d[cur.c][cur.s]) continue;
            visited.add(cur.c);

            for (Node next : g[cur.c]) {

                for (int i = 0; cur.s + i * C[cur.c] - next.a <= MAX; i++) {
//                for (int i = 0; i <= 75; i++) {
                    if (cur.s + i * C[cur.c] < next.a) continue;
//                    if (cur.s + i * C[cur.c] - next.a > MAX) break;
                    if (d[next.to][cur.s + i * C[cur.c] - next.a] > cur.v + next.b + i * D[cur.c]) {
                        d[next.to][cur.s + i * C[cur.c] - next.a] = cur.v + next.b + i * D[cur.c];
                        q.add(new Value(next.to, cur.v + next.b + i * D[cur.c], cur.s + i * C[cur.c] - next.a));
                    }
                }
            }
            if (visited.size() == n) break;
        }

        return d;
    }

    class Value implements Comparable<Value> {
        int c;
        long v;
        int s;

        Value(int c, long v, int s) {
            this.c = c;
            this.v = v;
            this.s = s;
        }

        @Override
        public int compareTo(Value o) {
            if (v != o.v)
                return Long.compare(v, o.v);
            return Long.compare(s, o.s);
        }
    }


    class Edge extends ArrayList<Node> {
    }

    class Node {
        int to, a;
        long b;

        Node(int to, int a, long b) {
            this.to = to;
            this.a = a;
            this.b = b;
        }
    }

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

    public static void main(String[] args) {
        new Main().run();
    }
}