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.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int M = scanner.nextInt();
    int S = scanner.nextInt();
    Node[] nodes = new Node[N];
    for (int i = 0; i < N; i++) nodes[i] = new Node();
    int[] A = new int[M];
    int[] B = new int[M];
    int[] E = new int[M];
    for (int i = 0; i < M; i++) {
      int u = scanner.nextInt() - 1;
      int v = scanner.nextInt() - 1;
      A[i] = scanner.nextInt();
      B[i] = scanner.nextInt();
      nodes[u].edges.add(i);
      nodes[v].edges.add(i);
      E[i] = u ^ v;
    }
    int[] C = new int[N];
    int[] D = new int[N];
    for (int i = 0; i < N; i++) {
      C[i] = scanner.nextInt();
      D[i] = scanner.nextInt();
    }

    long INF = 10_000_000_000_000L;
    int G = 2500;
    long[][] dp = new long[N][G + 1];
    for (int i = 0; i < N; i++) Arrays.fill(dp[i], INF);
    for (int i = 0; i <= S && i <= G; i++) dp[0][i] = 0;
    for (int k = 0; k < N + 1; k++) {
      for (int i = 0; i < N; i++) {
        for (int j = 0; j <= G; j++) {
          if (j >= C[i]) dp[i][j] = Math.min(dp[i][j], dp[i][j - C[i]] + D[i]);
          for (int e : nodes[i].edges)
            if (j + A[e] <= G) dp[i][j] = Math.min(dp[i][j], dp[E[e] ^ i][j + A[e]] + B[e]);
        }
      }
    }
    for (int i = 1; i < N; i++) {
      long min = INF;
      for (int j = 0; j <= 2500; j++) {
        min = Math.min(min, dp[i][j]);
      }
      System.out.println(min);
    }
  }

  private static class Node {
    private final List<Integer> edges = new ArrayList<>();
  }
}
