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 heappop, heappush

N, M, S = map(int, input().split())
X = [list(map(int, input().split())) for _ in range(M)]
Y = [list(map(int, input().split())) for _ in range(N)]

lim = 2501
graph = [[] for _ in range((N + 1) * lim)]

for u, v, a, b in X:
    for n in range(a, lim):
        graph[u * lim + n].append((v * lim + (n - a), b))
        graph[v * lim + n].append((u * lim + (n - a), b))

for n, (c, d) in enumerate(Y):
    for x in range(lim - 1):
        graph[(n + 1) * lim + x].append(
            ((n + 1) * lim + min(lim - 1, x + c), d))


INF = 10 ** 20
start = lim + min(S, lim - 1)
dist = [INF] * ((N + 1) * lim + 1)
dist[start] = 0
pq = []
heappush(pq, (0, start))

while pq:
    _, u = heappop(pq)
    for v, c in graph[u]:
        if dist[v] > dist[u] + c:
            dist[v] = dist[u] + c
            heappush(pq, (dist[v], v))

ans = []
for i in range(2, N + 1):
    ans.append(min(dist[i * lim:i * lim + lim]))

print(*ans, sep="\n")
