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
max_s = 2500

n,m,s = map(int, input().split())
s = min(s, max_s)
conv = lambda i,j : i*10000+j

G = [[] for i in range(conv(n-1, max_s)+1)]
for i in range(m):
    u,v,a,b = map(int, input().split())
    u,v = u-1, v-1
    for j in range(max_s+1):
        if j-a >= 0:
            G[conv(u,j)].append((conv(v,j-a), b))
            G[conv(v,j)].append((conv(u,j-a), b))

for i in range(n):
    c,d = map(int, input().split())
    for j in range(max_s+1):
        if j+c > max_s:
            break
        G[conv(i,j)].append((conv(i,j+c), d))

from heapq import heappush, heappop
def dijkstra(G, start):
    n = len(G)
    dist = [float("inf")]*n
    visited = set()
    heap = [(0, start)]

    while heap and len(visited) != n:
        d, cur = heappop(heap)
        if cur in visited:
            continue
        visited.add(cur)
        dist[cur] = d
        for to, cost in G[cur]:
            if to not in visited:
                heappush(heap, (d+cost, to))
    return dist

dist = dijkstra(G, conv(0, s))
for i in range(1,n):
    cost = min(dist[conv(i,0):conv(i,max_s)+1])
    print(cost)
