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
INF=1000000000000000 # 1e15
MAX=50*50 # 2500

n,m,s=map(int,input().split())
u,v,a,b,c,d=[0]*m,[0]*m,[0]*m,[0]*m,[0]*n,[0]*n
for i in range(m):
  u[i],v[i],a[i],b[i]=map(int,input().split())
  u[i] -= 1
  v[i] -= 1
for i in range(n):
  c[i],d[i]=map(int,input().split())

dp=[[INF] * (MAX + 1) for _ in range(n)]
#dp[i][j]=k <=> 最短k分後に街iに所持銀貨j枚で行ける
dp[0][min(MAX, s)] = 0
updated = True
while updated:
  updated = False
  for i in range(n):
    for j in range(MAX + 1):
      dest = min(MAX, j + c[i])
      if dp[i][j] + d[i] < dp[i][dest]:
        updated = True
        dp[i][dest] = dp[i][j] + d[i]
  for i in range(m):
    for j in range(a[i], MAX+1):
      # u[i] -> v[i]
      if dp[u[i]][j] + b[i] < dp[v[i]][j - a[i]]:
        updated = True
        dp[v[i]][j - a[i]] = dp[u[i]][j] + b[i]
      # v[i] -> u[i]
      if dp[v[i]][j] + b[i] < dp[u[i]][j - a[i]]:
        updated = True
        dp[u[i]][j - a[i]] = dp[v[i]][j] + b[i]
for i in range(1, n):
  print(min(dp[i]))
