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
input =sys.stdin.readline
import heapq

def dijkstra(edge, start):
    n = len(edge)
    dist = [float("inf")]*n
    dist[start] = 0
    que = [(0, start)]
    while que:
        d, v = heapq.heappop(que)
        if dist[v] < d:
            continue
        for nv, nd in edge[v]:
            if dist[nv] > d + nd:
                dist[nv] = d + nd
                heapq.heappush(que, (dist[nv], nv))
    return dist

n, m, s = map(int, input().split())
G = [[] for _ in range(n)]
if s >= 2500:
  for _ in range(m):
    u, v, a, b = map(int, input().split())
    u -= 1
    v -= 1
    G[u].append((v, b))
    G[v].append((u, b))
  L = dijkstra(G, 0)
else:
  for _ in range(m):
    u, v, a, b = map(int, input().split())
    u -= 1
    v -= 1
    G[u].append((v, a, b))
    G[v].append((u, a, b))
  CD = [list(map(int, input().split())) for _ in range(n)]
  edge = [[] for _ in range(n*2500)]
  for i in range(n):
    c, d = CD[i]
    for k in range(2500):
      for v, a, b in G[i]:
        if k >= a:
          edge[2500*i+k].append((2500*v+(k-a), b))
      if k+c < 2500:
        edge[2500*i+k].append((2500*i+(k+c), d))
  dist = dijkstra(edge, s)
  L = [0]*n
  for i in range(n):
    L[i] = min(dist[2500*i:2500*(i+1)])
print(*L[1:], sep="\n")