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
def dijkstra_heap(s):
  d=[float("inf")]*n
  used=[True]*n 
  d[s]=0
  used[s]=False
  edgelist=[]
  for e in edge[s]:
    heapq.heappush(edgelist,e)
  while len(edgelist):
    minedge=heapq.heappop(edgelist)
    if not used[minedge[1]]:
      continue
    v=minedge[1]
    d[v]=minedge[0]
    used[v]=False
    for e in edge[v]:
      if used[e[1]]:
        heapq.heappush(edgelist,[e[0]+d[v],e[1]])
  return d

N,m,s=map(int,input().split())
L=[list(map(int,input().split())) for i in range(m)]
L2=[list(map(int,input().split())) for i in range(N)]
n=N*(50*(N-1)+1)
edge=[[] for i in range(n)]
for i in range(m):
  for j in range(50*(N-1)+1):
    if j>=L[i][2]:
      edge[L[i][0]-1+N*j].append([L[i][3],L[i][1]-1+N*(j-L[i][2])])
      edge[L[i][1]-1+N*j].append([L[i][3],L[i][0]-1+N*(j-L[i][2])])
for i in range(N):
  for j in range(50*(N-1)):
    edge[i+N*j].append([L2[i][1],i+N*min(50*(N-1),j+L2[i][0])])
ans=[[] for i in range(N)]
D=dijkstra_heap(min(50*(N-1),s)*N)
for i in range(len(D)):
  ans[i%N].append(D[i])
for i in range(1,N):
  print(min(ans[i]))