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 heapq
import sys

input = sys.stdin.readline

inf = 10**13

N,M,S = map(int,input().split())

data = [[] for _ in range(N)]

sumA = 0

for i in range(M):
    U,V,A,B = map(int,input().split())
    sumA += A
    U -= 1
    V -= 1
    data[U].append((B,V,A))###(距離, 行き先, 費用)
    data[V].append((B,U,A))###(距離, 行き先, 費用)

P = sumA

C = [0]*N
D = [0]*N

for i in range(N):
    C[i],D[i] = map(int,input().split())

dist = [[inf]*(P+1) for _ in range(N)]
#visit = [[0]*(P+1) for _ in range(N)]
###heapにぶち込んだことがあるかないか

q = [(0,0,min(S,P))]###(仮の最短距離, 頂点番号, 所持金)
heapq.heapify(q)

while q:
    v = heapq.heappop(q)
    if v[0] >= dist[v[1]][v[2]]:
        continue
    dist[v[1]][v[2]] = v[0]

    if v[2] + C[v[1]] <= P:
        if dist[v[1]][v[2]+C[v[1]]] == inf:
            heapq.heappush(q,(v[0]+D[v[1]],v[1],v[2]+C[v[1]]))

    for x in data[v[1]]:
        if x[2] > v[2]:
            continue
        if dist[x[1]][v[2]-x[2]] == inf:
            heapq.heappush(q,(v[0]+x[0],x[1],v[2]-x[2]))
    
    ###
    #print(P)
    #print(q)
    #print(v)
    ###

for i in range(1,N):
    print(min(dist[i]))
        

