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.*;
public class Main {

	static long[][] dp;
	static PriorityQueue<S> q;
	
	static class S{
		int a, b;
		long c;
		S(int a, int b, long c){
			this.a = a;
			this.b = b;
			this.c = c;
		}
	}
	
	static void push(int a, int b, long c) {
		if(b < 0)
			return;
		if(dp[a][b] <= c)
			return;
		dp[a][b] = c;
		q.add(new S(a, b, c));
	}
	
	public static void main(String[] args) {
		long INF = Long.MAX_VALUE/10;
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		int S = sc.nextInt();
		int smax = 50 * N;
		Set<S>[] E = new Set[N];
		for(int i = 0; i < N; i++)
			E[i] = new HashSet<>();
		for(int i = 0; i < M; i++) {
			int U = sc.nextInt()-1;
			int V = sc.nextInt()-1;
			int A = sc.nextInt();
			int B = sc.nextInt();
			E[U].add(new S(V, A, B));
			E[V].add(new S(U, A, B));
		}
		int[] C = new int[N];
		int[] D = new int[N];
		for(int i = 0; i < N; i++) {
			C[i] = sc.nextInt();
			D[i] = sc.nextInt();
		}
		dp = new long[N][smax+1];
		for(int i = 0; i < N; i++)
			Arrays.fill(dp[i], INF);
		dp[0][Math.min(smax, S)] = 0;
		q = new PriorityQueue<>((p1,p2)->Long.compare(p1.c,p2.c));
		q.add(new S(0, Math.min(smax, S), 0));
		while(q.size() > 0) {
			S s = q.poll();
			if(dp[s.a][s.b] != s.c)
				continue;
			push(s.a, Math.min(s.b + C[s.a], smax), s.c+D[s.a]);
			for(S e : E[s.a])
				push(e.a, s.b-e.b, s.c+e.c);
		}
		for(int i = 1; i < N; i++) {
			long ans = INF;
			for(long t : dp[i])
				ans = Math.min(ans, t);
			System.out.println(ans);
		}
	}

}