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


dp = {}
hq = []

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

MAX_S = (N-1)*50
INF = 10**18
S = min(S, MAX_S)

for i in range(1,N+1):
    dp[i] = {}
    for j in range(MAX_S+1):
        dp[i][j] = INF

uv = {}
for i in range(N):
    uv[i+1] = []
for i in range(M):
    U, V, A, B = map(int, input().split())
    uv[U].append((V,A,B))
    uv[V].append((U,A,B))

c = {}
d = {}
for i in range(N):
    C, D = map(int, input().split())
    c[i+1] = C
    d[i+1] = D

def q_push(v, money, time):
    if money < 0:
        return 
    if dp[v][money] <= time:
        return
    dp[v][money] = time
    heapq.heappush(hq, (time, v, money))


q_push(1,S,0)
while hq:
    now_t, now_v, now_m = heapq.heappop(hq)

    if dp[now_v][now_m] != now_t:
        continue

    new_money = min(now_m+c[now_v],MAX_S)
    q_push(now_v,new_money,now_t+d[now_v])
    for next_v, next_a, next_b in uv[now_v]:
        q_push(next_v,now_m-next_a,now_t+next_b)

for i in range(2, N+1):
    ans = INF
    for j in range(0, MAX_S+1):
        ans = min(ans, dp[i][j])
    
    print(ans)