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 heapq
n, m, s = map(int, input().split())
s = min(s, 2499)
UV = [[] for _ in range(n)]
for i in range(m):
    u, v, a, b = map(int, input().split())
    u -= 1
    v -= 1
    UV[u].append((v, a, b))
    UV[v].append((u, a, b))
CD = [tuple(map(int, input().split()))for _ in range(n)]
MAXS = 2500
NV = MAXS*n
edges = [[] for _ in range(NV)]


def to_v(i, j):
    return MAXS*i+j

for i, (c, d) in enumerate(CD):
    for j in range(MAXS-c):
        edges[to_v(i, j)].append((d, to_v(i, j)+c))

for i in range(n):
    for i2, a, b in UV[i]:
        for j in range(MAXS-a):
            edges[to_v(i, a+j)].append((b, to_v(i2, j)))


def dijkstra(init_v):
    next_v = [(0, init_v)]
    INF = 10**18
    dist = [INF]*NV
    dist[init_v] = 0
    while next_v:
        d, v = heapq.heappop(next_v)
        if dist[v] < d:
            continue
        for d, v2 in edges[v]:
            if dist[v2] <= dist[v]+d:
                continue
            dist[v2] = dist[v]+d
            heapq.heappush(next_v, (dist[v2], v2))
    return dist


INF = 10**18
dist = dijkstra(s)
for i in range(1, n):
    ans = INF
    for j in range(MAXS):
        ans = min(ans, dist[to_v(i, j)])
    print(ans)