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
inf = 10**20
h = 2500
n, m, s = map(int, input().split())
s = min(h-1, s)
# A[i][j]=iからjへのコスト
A = [[inf for j in range(n)] for i in range(n)]
# B[i][j]=iからjへの時間
B = [[inf for j in range(n)] for i in range(n)]
for _ in range(m):
    u, v, a, b = map(int, input().split())
    A[u-1][v-1] = a
    A[v-1][u-1] = a
    B[u-1][v-1] = b
    B[v-1][u-1] = b
CD = [tuple(map(int, input().split())) for i in range(n)]
C = [c for c, d in CD]
D = [d for c, d in CD]
# G[i,x][j,y]=状態(i,x)から(j,y)に変わるのにかかる時間
G = {(i, x): {} for i in range(n) for x in range(h)}
for i in range(n):
    for j in range(n):
        for x in range(h):
            if i == j:
                G[i, x][i, x] = 0
                if x-1 >= 0:
                    G[i, x][i, x-1] = 0
                if x+C[i] < h:
                    G[i, x][i, x+C[i]] = D[i]
            elif x >= A[i][j]:
                G[i, x][j, x-A[i][j]] = B[i][j]
# T[i,x]=状態(0,s)から(i,x)にかかる時間
T = {(i, x): inf for i in range(n) for x in range(h)}
T[0, s] = 0
Q = [(0, (0, s))]
heapq.heapify(Q)
while Q:
    _, (i, x) = heapq.heappop(Q)
    for j, y in G[i, x].keys():
        if T[i, x]+G[i, x][j, y] < T[j, y]:
            T[j, y] = T[i, x] + G[i, x][j, y]
            heapq.heappush(Q, (T[j, y], (j, y)))
for i in range(1, n):
    print(T[i, 0])
