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
import sys

sys.setrecursionlimit(10 ** 7)

read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N, M, S = map(int, input().split())
S = min(2500, S)

# INPUT
PATH = [[] for _ in range(N + 1)]
CHANGES = [None] * (N + 1)

for _ in range(M):
    U, V, A, B = map(int, input().split())
    PATH[U].append((V, A, B))
    PATH[V].append((U, A, B))

for i in range(N):
    C, D = map(int, input().split())
    CHANGES[i + 1] = (C, D)

Q = []
INF = float("inf")
TIME = [[INF] * 2501 for _ in range(N + 1)]
USED = [[False] * 2501 for _ in range(N + 1)]

Q.append((0, 1, S))
TIME[1][S] = 0

while Q:
    t, u, g = heappop(Q)

    # そこで銀貨を増やす
    # 2500枚未満の場合
    if g < 2500:
        c, d = CHANGES[u]
        ng = min(g + c, 2500)
        nt = t + d

        if nt < TIME[u][ng]:
            TIME[u][ng] = nt
            heappush(Q, (nt, u, ng))

    # 次の頂点に進む
    for v, a, b in PATH[u]:
        if g < a:
            continue

        ng = g - a
        nt = t + b
        if nt < TIME[v][ng]:
            TIME[v][ng] = nt
            heappush(Q, (nt, v, ng))

ANS = [min(x) for x in TIME[2:]]
print("\n".join(map(str, ANS)))
