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 heappop, heappush

n,m,s = map(int,input().split())
e = [[] for _ in range(n)]
for i in range(m):
  u,v,a,b = map(int,input().split())
  u -= 1
  v -= 1
  e[u].append((v,a,b))
  e[v].append((u,a,b))
c = []
d = []
for i in range(n):
  x,y = map(int,input().split())
  c.append(x)
  d.append(y)

INF = float('inf')
def dijkstra(s,n,g): # sからの最短距離 頂点数n g[v]=[(u1,w1),(u2,w2)....]:vからuへの有向辺があり、重みw
  que = []
  d = [INF for _ in range(n)]
  d[s] = 0
  heappush(que,(0,s))
  while len(que) > 0:
    p = heappop(que)
    v = p[1]
    if d[v] < p[0]:
      continue
    for u,w in g[v]:
      if d[u] > d[v]+w:
        d[u] = d[v]+w
        heappush(que,(d[u],u))
  return d

def newv(v,s):
  return v*2500+s

s = min(s,2499)
e2 = [[] for _ in range(n*2500)]
for u in range(n):
  for j in range(2500):
    #if j+c[u] < 2500:
    #  e2[newv(u,j)].append((newv(u,j+c[u]),d[u]))
    e2[newv(u,j)].append((newv(u,min(j+c[u],2499)),d[u]))
    for v,a,b in e[u]:
      if j >= a:
        e2[newv(u,j)].append((newv(v,j-a),b))

l = dijkstra(newv(0,s),n*2500,e2)
for i in range(1,n):
  ans = INF
  for j in range(2500):
    ans = min(l[newv(i,j)],ans)
  print(ans)
