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
#include <stdio.h>
#include <algorithm>
using namespace std;

const int coin = 2555;
long long t[55][coin + 1];
int N, M, S, X[111], Y[111], A[111], B[111], C[55], D[55];

int main()
{
	scanf ("%d %d %d", &N, &M, &S);
	if (S > coin) S = coin;

	for (int i = 1; i <= M; i++) scanf ("%d %d %d %d", &X[i], &Y[i], &A[i], &B[i]);
	for (int i = 1; i <= N; i++) scanf ("%d %d", &C[i], &D[i]);

	for (int i = 1; i <= N; i++) for (int j = 0; j <= coin; j++) t[i][j] = 1e18;
	t[1][S] = 0;

	for (int k = 0; k < N * N; k++){
		for (int i = 1; i <= N; i++) for (int j = 0; j <= coin; j++){
			int k = j + C[i];
			if (k > coin) k = coin;
			t[i][k] = min(t[i][k], t[i][j] + D[i]);
		}

		for (int i = 1; i <= M; i++){
			for (int j = A[i]; j <= coin; j++){
				int k = j - A[i];
				t[X[i]][k] = min(t[X[i]][k], t[Y[i]][j] + B[i]);
				t[Y[i]][k] = min(t[Y[i]][k], t[X[i]][j] + B[i]);
			}
		}
	}

	for (int i = 2; i <= N; i++){
		long long ans = 1e18;
		for (int j = 0; j <= coin; j++) ans = min(ans, t[i][j]);
		printf ("%lld\n", ans);
	}

	return 0;
}