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 sys import stdin, setrecursionlimit
input = stdin.buffer.readline
from heapq import heappush, heappop

INF = 10 ** 18

N, M, S = map(int, input().split())
UVAB = [list(map(int, input().split())) for _ in range(M)]
CD = [list(map(int, input().split())) for _ in range(N)]

sumA = 0
for u, v, a, b in UVAB:
    sumA += a

def dijkstra(start, graph) -> list:
    dist = [[INF for _ in range(N + 1)] for _ in range(sumA + 1)]
    dist[start][1] = 0
    heap = []
    heappush(heap, (0, (start, 1)))
    while heap:
        v, (i, j) = heappop(heap)
        for w, (ni, nj) in graph[i][j]:
            if dist[ni][nj] > v + w:
                dist[ni][nj] = v + w
                heappush(heap, (dist[ni][nj], (ni, nj)))
    return dist

graph = [[[] for _ in range(N + 1)] for _ in range(sumA + 1)]

for u, v, a, b in UVAB:
    for i in range(a, sumA + 1):
        graph[i][u].append((b, (i - a, v)))
        graph[i][v].append((b, (i - a, u)))

for j in range(1, N + 1):
    c, d = CD[j - 1]
    for i in range(sumA):
        graph[i][j].append((d, (min(i + c, sumA), j)))

dist = dijkstra(min(S, sumA), graph)

for j in range(2, N + 1):
    answer = INF
    for i in range(sumA + 1):
        answer = min(dist[i][j], answer)
    print(answer)