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
# coding: utf-8
import sys
from heapq import heapify, heappop, heappush

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

# 銀貨を何枚持っているかの状態数、N*2501の都市, dijkstra
N, M, S = lr()
limit = 2500
S = min(S, limit)
graph = [[] for _ in range((N+1) * (limit+1))]  # 1-indexed
for _ in range(M):
    u, v, a, b = lr()
    for x in range(a, limit+1):
        graph[u*(limit+1) + x].append(((v*(limit+1)+x-a), b))
        graph[v*(limit+1) + x].append(((u*(limit+1)+x-a), b))

for i in range(N):
    i += 1
    c, d = lr()
    for x in range(limit-c+1):
        graph[i*(limit+1) + x].append((i*(limit+1) + x + c, d))

def dijkstra(start):
    INF = 10 ** 15
    dist = [INF] * ((N+1) * (limit+1))
    dist[start] = 0
    que = [(0, start)]
    while que:
        d, prev = heappop(que)
        if dist[prev] < d:
            continue
        for next, time in graph[prev]:
            d1 = d + time
            if dist[next] > d1:
                dist[next] = d1
                heappush(que, (d1, next))
    return dist

dist = dijkstra(1*(limit+1)+S)
for i in range(2, N+1):
    answer = min(dist[i*(limit+1):(i+1)*(limit+1)])
    print(answer)
