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
N,M,S=map(int,input().split())
from heapq import heappop,heappush 

inf = float('inf')
cost=[[] for _ in range(N)]
exc=[]


for _ in range(M):
    u,v,a,b=map(int,input().split())
    u,v=u-1,v-1
    cost[u].append((v,a,b))
    cost[v].append((u,a,b))
for _ in range(N):
    c,d=map(int,input().split())
    exc.append((c,d))

di=[inf]*2451*N
if S>2450:
    S=2450
di[S]=0
hq=[(0,S)] #時間,都市名*2451+銀貨

while hq:
    t,v=heappop(hq)
    if t>di[v]:
        continue
    u=v//2451 #都市名
    s=v%2451 #所持金

    for x,a,b in cost[u]:
        if s>=a:
            v_new=x*2451+s-a
            t_new=t+b
            if t_new<di[v_new]:
                di[v_new]=t_new
                heappush(hq,(t_new,v_new))
    
    c,d=exc[u]
    if s+c>2450:
        c=2450-s
    v_new=v+c
    t_new=t+d
    if t_new<di[v_new]:
        di[v_new]=t_new
        heappush(hq,(t_new,v_new))

for i in range(1,N):
    m=di[i*2451]
    for j in range(1,2451):
        m=min(m,di[i*2451+j])
    print(m)