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
n, m, s = map(int, input().split())

edges = [[] for _ in range(n)]
amax = 0
for _ in range(m):
   u, v, a, b = map(int, input().split())
   edges[u-1].append((v-1, a, b))
   edges[v-1].append((u-1, a, b))
   amax = max(a, amax)

lcoins = amax * (n - 1)

cost = []
time = []
for _ in range(n):
   c, d = map(int, input().split())
   cost.append(c)
   time.append(d)

from heapq import *


INF = 1<<61
M = 2500
if s >= M:
   s = M - 1

dist = [[INF] * M for _ in range(n)]
dist[0][s] = 0
q = [(0, s, 0)]

while q:
   dv, num, v = heappop(q)

   if dist[v][num] < dv:
      continue

   # update
   if num + cost[v] < M and dv + time[v] < dist[v][num+cost[v]]:
      dist[v][num+cost[v]] = dv + time[v]
      heappush(q, (dv+time[v], num+cost[v], v))

   for next, a, b in edges[v]:
      if num >= a and dv+b < dist[next][num-a]:
         dist[next][num-a] = dv+b
         heappush(q, (dv+b, num-a, next))


for i in range(1, n):
   print(min(dist[i]))
