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
from heapq import heappush,heappop

N,M,S=map(int,input().split())
edges = [tuple(map(int,input().split())) for _ in range(M)]
cd = [tuple(map(int,input().split())) for _ in range(N)]
INF= 10**15
mod=55

mmax = sum([edges[i][2] for i in range(M)])
if S>mmax:
  S = mmax

G=[{} for _ in range((mmax+1)*mod+N+1)]
for i in range(M):
  u,v,a,b=edges[i]
  for p in range(a,mmax+1):
    G[u+p*mod][v+(p-a)*mod] = b
    G[v+p*mod][u+(p-a)*mod] = b
    
for i in range(N):
  c,d=cd[i]
  for p in range(mmax+1-c):
    G[i+1+p*mod][i+1+(p+c)*mod] = d
    
def dijkstra(start = 0):
    
    d = [INF for i in range(len(G)+1)]
    d[start] = 0
    
    que = []
    heappush(que, (0, start))

    while len(que):
        p = heappop(que)
        v = p[1]
        
        if d[v] < p[0]:
            continue
            
        for u in G[v].keys():
            if d[u] > d[v] + G[v][u]:
                d[u] = d[v] + G[v][u]
                heappush(que, (d[u], u))
    
    return d
  
start = 1+S*mod
d = dijkstra(start)

ans=[INF]*(N+1)
for i in range(2,len(G)+1):
  v = i%mod
  if v>N or v<2:continue
  ans[v] = min(ans[v],d[i])
  
print(*ans[2:])