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()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")


n,m,s = list(map(int, input().split()))
cd = [None]*n
from collections import defaultdict
ns = defaultdict(set)
nns = defaultdict(set)
uvab = [None]*m
for i in range(m):
    u,v,a,b = list(map(int, input().split()))
    u -= 1
    v -= 1
    uvab[i] = (u,v,a,b)
    ns[u].add((a,b,v))
    ns[v].add((a,b,u))
for i in range(n):
    cd[i] = list(map(int, input().split()))

ms = max(item[2] for item in uvab)*(n-1)
for u in range(n):
    for ss in range(ms+1):
        for a,b,v in ns[u]:
            if ss>=a:
                nns[u,ss].add((b, (v,ss-a)))
        c,d = cd[u]
        if ss+c<=ms:
            nns[u,ss].add((d, (u,ss+c)))
ns = nns
def dijkstra(start):
    import heapq
    vals = {}
    h = [(0, start)] # (距離, ノード番号)
    vals[start] = 0
    while h:
        val, u = heapq.heappop(h)
        for d, v in ns[u]:
            if v not in vals or vals[v]>val+d:
                vals[v] = val+d
                heapq.heappush(h, (vals[v], v))
    return vals
s = min(s, ms)
vals = dijkstra((0,s))
ans = [None]*(n-1)
for u in range(1,n):
    ans[u-1] = min(vals[u,s] for s in range(ms+1))
write("\n".join(map(str, ans)))