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()

import heapq
from collections import defaultdict
h = []
heapq.heapify(h)

n, m, s = list(map(int, input().split()))
a, b = [[None] * m for _ in range(2)]
c, d = [[None] * n for _ in range(2)]

ns = defaultdict(list)
uv2e = {}
for i in range(m):
    u,v,a[i],b[i] = list(map(int, input().split()))
    ns[u-1].append(v-1)
    ns[v-1].append(u-1)
    uv2e[u-1,v-1] = i
    uv2e[v-1,u-1] = i
for i in range(n):
    c[i], d[i] = list(map(int, input().split()))

maxs = max(a)*(n-1)
s = min(s, maxs)
state = (0, 0, s) # (距離, ノード, 銀貨枚数)
ans = [[None] * (maxs+1) for _ in range(n)]
heapq.heappush(h, state)
count = 0
while h:
    count += 1
    dd, u, ss = heapq.heappop(h)
#     print(dd, u, ss)
    if ans[u][ss] is not None and ans[u][ss] <= dd:
        continue
    ans[u][ss] = dd
    
    # グラフの隣接点
    for v in ns[u]:
        e= uv2e[u,v]
        if ss<a[e] or ans[v][ss-a[e]] is not None:
            continue
        heapq.heappush(h, (dd + b[e], v, ss-a[e]))
    # 両替
    if ss+c[u] <= maxs:
        heapq.heappush(h, (dd + d[u], u, ss+c[u]))
    elif ss < maxs:
        heapq.heappush(h, (dd + d[u], u, maxs))

for item in ans[1:]:
    print(min(item))
