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 math
import heapq

n,m,s = map(int,input().split())

road = []
for i in range(m):
    u,v,a,b = map(int,input().split())
    road.append([u,v,a,b])

change = []
for i in range(n):
    c,d = map(int,input().split())
    change.append([c,d])

tree = [[] for _ in range(n+1)]

for i in range(m):  
    r = road[i]
    tree[r[0]].append([r[0],r[1],r[2],r[3]])
    tree[r[1]].append([r[1],r[0],r[2],r[3]])

least_cost = [[] for _ in range(n+1)]
for l in range(n+1):
    least_cost[l] = [math.inf]*2501

queue = []

#hwd [location, money, time]
def dijakstra(q):
    while q != []:
        hwd = heapq.heappop(q)

        if least_cost[hwd[1]][hwd[2]] > hwd[0]:
            least_cost[hwd[1]][hwd[2]] = hwd[0]
            
            for node in tree[hwd[1]]:
                # from node, to node , money, time
                if hwd[2] - node[2] >= 0:
                    heapq.heappush(q,(hwd[0]+node[3],node[1],hwd[2]-node[2]))
            if hwd[2] + change[hwd[1]-1][0] <= 2500:
                heapq.heappush(q,(hwd[0]+change[hwd[1]-1][1],hwd[1],hwd[2]+change[hwd[1]-1][0]))
    return least_cost

if s > 2500:
    s = 2500
queue.append((0,1,s))

ans = dijakstra(queue)

for i in range(2,n+1):
    print(min(ans[i]))