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
from collections import defaultdict
import heapq
N,M,S = map(int,input().split())
E=[list(map(int,input().split())) for _ in range(M)]

C=[list(map(int,input().split())) for _ in range(N)]

A_max=max([a for _,_,a,_ in E])*N

D=defaultdict(lambda :[])

for u,v,a,b in E:
    u-=1;v-=1
    for from_a in range(A_max+1):
        if from_a-a<0:
            continue
        from_state_u=(u,from_a)
        to_state_v=(v,from_a-a)
        D[from_state_u].append((b,to_state_v))

        from_state_v=(v,from_a)
        to_state_u=(u,from_a-a)
        D[from_state_v].append((b,to_state_u))

for n,[c,d] in enumerate(C):
    for a in range(A_max+1):
        a_after=min(A_max,a+c)
        from_state=(n,a)
        to_state=(n,a_after)
        D[from_state].append((d,to_state))


start=(0,min(A_max,S))
Q=[(0,start)]

Dist=defaultdict(lambda :float('inf'))

Dist[start]=0

while Q:
    q = heapq.heappop(Q)
    cur_time,cur_state=q
    if Dist[cur_state]<cur_time:
        continue
    for cost,next_state in D[cur_state]:
        next_time = cost+cur_time

        if Dist[next_state]>next_time:
            Dist[next_state]=next_time
            heapq.heappush(Q,(next_time,next_state))

for dest in range(1,N):
    ans=float("inf")
    for a in range(A_max+1):
        state=(dest,a)
        ans=min(ans,Dist[state])
    print(ans)
