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 sys
from heapq import *
I=sys.stdin.readlines()
N,M,S=map(int,I[0].split())
S=min(S,2499)
G=[[] for i in range(2500*N)]
u,v,a,b=0,0,0,0
for i in range(M):
  u,v,a,b=map(int,I[i+1].split())
  u,v=u-1,v-1
  for j in range(a,2500):
    G[u+j*N].append((v+(j-a)*N,b))
    G[v+j*N].append((u+(j-a)*N,b))
for i in range(N):
  a,b=map(int,I[M+i+1].split())
  for j in range(a,2500):
    G[i+(j-a)*N].append((i+j*N,b))
D=[0]*(2500*N)
INF=10**15
def ijk(s):
  global D,INF,G
  for i in range(len(G)):
    D[i]=INF
  D[s]=0
  Q=[]
  heapify(Q)
  heappush(Q,(0,s))
  p,v,e=0,0,0
  while len(Q):
    p=heappop(Q)
    v=p[1]
    if D[v]<p[0]:
      continue
    for i in range(len(G[v])):
      e=G[v][i]
      if D[e[0]]>D[v]+e[1]:
        D[e[0]]=D[v]+e[1]
        heappush(Q,(D[e[0]],e[0]))

ijk(S*N)
for i in range(1,N):
  print(min(D[i::N]))