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 heapq import heapify, heappush as hpush, heappop as hpop
N, M, S = map(int, input().split())
k = 2451
X = [[] for _ in range(N * k)]
for _ in range(M):
    u, v, a, b = map(int, input().split())
    u, v = u-1, v-1
    for i in range(a, k):
        X[u * k + i].append([v * k + i - a, b])
        X[v * k + i].append([u * k + i - a, b])

for i in range(N):
    c, d = map(int, input().split())
    for j in range(k - 1):
        X[i * k + j].append([i * k + min(j + c, k - 1), d])

def dijkstra(n, E, i0=0):
    h = [[0, i0]]
    D = [-1] * n
    done = [0] * n
    D[i0] = 0
    
    while h:
        d, i = hpop(h)
        done[i] = 1
        for j, w in E[i]:
            nd = d + w
            if D[j] < 0 or D[j] > nd:
                if done[j] == 0:
                    hpush(h, [nd, j])
                    D[j] = nd
    return D

D = dijkstra(N * k, X, min(S, k - 1))
print(*[min(D[i * k:(i+1) * k]) for i in range(N)][1:], sep = "\n")