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())
UVAB=[list(map(int,input().split())) for i in range(M)]
C,D=zip(*[list(map(int,input().split())) for i in range(N)])
c=[[] for i in range(N)]
A=[[] for i in range(N)]
B=[[] for i in range(N)]
for u,v,a,b in UVAB:
    c[u-1].append(v-1)
    c[v-1].append(u-1)
    A[u-1].append(a)
    A[v-1].append(a)
    B[u-1].append(b)
    B[v-1].append(b)
inf=10**20
from heapq import heappop,heappush
X=50*(N-1)
def dij(p,s):
    d=[[inf]*(X+1) for i in range(N)]
    d[p][s]=0
    q=[]
    heappush(q,(0,(p,s)))
    while q:
        t,(p,s)=heappop(q)
        for i in range(1,(X+1-s)//C[p]):
            if X<s+i*C[p] or d[p][s+i*C[p]]<=d[p][s]+i*D[p]:break
            d[p][s+i*C[p]]=d[p][s]+i*D[p]
            heappush(q,(d[p][s]+i*D[p],(p,s+i*C[p])))
        for j,n in enumerate(c[p]):
            if 0<=s-A[p][j] and d[n][s-A[p][j]]>d[p][s]+B[p][j]:
                d[n][s-A[p][j]]=d[p][s]+B[p][j]
                heappush(q,(d[n][s-A[p][j]],(n,s-A[p][j])))
    return d
d=dij(0,min(S,X))
print(*[min(x) for x in d[1:]],sep='\n')