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 heapq
N,M,S = map(int,input().split())
MAX_COST = 2500
S = min(S, MAX_COST)

G = [ [] for _ in range(N) ]
for i in range(M):
  u,v,s,t = map(int,input().split())
  u, v = u-1, v-1
  G[u].append([v, s, t])
  G[v].append([u, s, t])

change_rate, change_cost = [], []
for i in range(N):
  rate, cost = map(int,input().split())
  change_rate.append(rate)
  change_cost.append(cost)

# dp[i][silver] = 頂点iにいて、銀貨をsilver枚持つときの最小時間
dp = [ [float('inf')]*(MAX_COST+1) for _ in range(N) ]
dp[0][S] = 0

# time,node,silver
pq = [(0,0,S)]
while pq:
  time,node,silver = heapq.heappop(pq)

  # going through exchange station
  next_silver = min(silver + change_rate[node], MAX_COST)
  next_time = time + change_cost[node]
  if next_time < dp[node][next_silver]:
    dp[node][next_silver] = next_time
    heapq.heappush(pq, (next_time, node, next_silver))

  for nei, ns, nt in G[node]:
    remain_silver = min(silver - ns, MAX_COST)
    if remain_silver < 0: continue

    next_time = time + nt
    if dp[nei][remain_silver] > next_time:
      dp[nei][remain_silver] = next_time
      heapq.heappush(pq, (next_time, nei, remain_silver))

for d in dp[1:]:
  print(min(d))