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 sys
input = lambda: sys.stdin.readline().rstrip()

n, m, s = map(int, input().split())
s = min(s, 2500)
from collections import defaultdict
# O(ElogV)
import heapq
def dijkstra(s, x):
    # 始点sから各頂点への最短距離
    d = defaultdict(lambda: float("inf"))
    d[(s, x)] = 0
    # 各頂点が訪問済みかどうか
    used = defaultdict(lambda: False)
    used[(s, x)] = True
    # 仮の距離を記録するヒープ
    que = []
    for e in edge[(s, x)]:
        heapq.heappush(que, e)
    while que:
        minedge = heapq.heappop(que)
        if used[minedge[1]]:
            continue
        v = minedge[1]
        d[v] = minedge[0]
        used[v] = True
        for e in edge[v]:
            if not used[e[1]]:
                heapq.heappush(que, (e[0] + d[v], e[1]))
    return d

max_money = 2500
edge = defaultdict(list)
for _ in range(m):
    u, v, a, b = map(int, input().split())
    u -= 1
    v -= 1
    for i in range(a, max_money + 1):
        edge[(u, i)].append((b, (v, i - a)))
        edge[(v, i)].append((b, (u, i - a)))
for j in range(n):
    c, d = map(int, input().split())
    for i in range(max_money):
        if i + c <= max_money:
            edge[(j, i)].append((d, (j, i + c)))

ans = [10**18]*n

d = dijkstra(0, s)
for i in d:
    x, y = i
    cost = d[i]
    ans[x] = min(ans[x], cost)
for i in ans[1:]:
    print(i)
