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
from heapq import *
import sys
input = sys.stdin.buffer.readline
n, m, s = map(int, input().split())
lim = 2510
di = [[float("inf")] * lim for _ in range(n)]
di[0][min(s,lim-1)] = 0
con = [[] for _ in range(n)]
for _ in range(m):
	u, v, a, b = map(int, input().split())
	con[u - 1].append((v - 1, a, b))
	con[v - 1].append((u - 1, a, b))
q = [(0, min(s,lim-1), 0)]
cd = [list(map(int, input().split())) for _ in range(n)]
while q:
	time, have, cur = heappop(q)
	if di[cur][have] < time:
		continue
	c, d = cd[cur]
	if have + c < lim and di[cur][have + c] > time + d:
		di[cur][have + c] = time + d
		heappush(q, (time + d, have + c, cur))
	for nxt, dh, dt in con[cur]:
		if have >= dh and di[nxt][have - dh] > time + dt:
			di[nxt][have - dh] = time + dt
			heappush(q, (time + dt, have - dh, nxt))
for p in di[1:]:
	print(min(p))
