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())

ways=[[] for _ in range(N)]
CD=[[0]*2 for _ in range(N)]

class Way:
  def __init__(self,t,f,ti):
    self.to=t
    self.fare=f
    self.time=ti 

for i in range(M):
  U,V,A,B=map(int,input().split())
  U-=1
  V-=1
  ways[U].append(Way(V,A,B))
  ways[V].append(Way(U,A,B))

for i in range(N):
  C,D=map(int,input().split())
  CD[i][0]=C
  CD[i][1]=D
  
#dp[i][j]:駅iにおいて銀貨をj枚所持するための最短時間
dp=[[float("inf")]*(50*N+1) for _ in range(N)]
S=min(S,50*N)
dp[0][S]=0
import heapq

#data型は(時間、駅、お金)
#q=heapq.heapify([(0,0,S)])
#q=heapq.heapify([])
q=[]
heapq.heappush(q,(0,0,S))
#print(q)
def Push(time,station,money):
  #print("called")
  global q,dp
  if money<0:
    return
  money=min(money,50*N)
  if dp[station][money] <= time:
    return
  dp[station][money] = time
  heapq.heappush(q,(time,station,money))

while q:
  #print("called!")
  dat = heapq.heappop(q)
  time,station,money=dat[0],dat[1],dat[2]
  
  if dp[station][money]!=time:
    continue
  #まず、両替。
  Push(time+CD[station][1],station,money+CD[station][0])
  #次に、各道について移動
  for w in ways[station]:
    Push(time+w.time,w.to,money-w.fare)

for i in range(1,N):
  ans=float("inf")
  for j in range(50*N+1):
    ans=min(ans,dp[i][j])
  print(ans)
    

  
  