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 sys
input = sys.stdin.readline
def main():
  n,m,s=map(int,input().split())
  uvab=[list(map(int,input().split())) for _ in range(m)]
  cd=[list(map(int,input().split())) for _ in range(n)]
  g=[[]for _ in range(n)]
  for u,v,a,b in uvab:
    g[u-1].append((v-1,a,b))
    g[v-1].append((u-1,a,b))
  import heapq
  inf=pow(10,21)
  max_s=50*50
  dp=[[inf]*(max_s+1)for _ in range(n)]
  kakutei=[[0]*(max_s+1)for _ in range(n)]
  dp[0][min(s,max_s)]=0
  todo=[[0,0,min(s,max_s)]] # [dp[v][i],v,i]
  heapq.heapify(todo)
  flgs=[0]*n
  flg=0
  ans=[inf]*n
  while flg<n:
    _,v,i=heapq.heappop(todo)
    if kakutei[v][i]==1:
      continue
    kakutei[v][i]=1
    if flgs[v]==0:
      flg+=1
      flgs[v]=1
    c,d=cd[v]
    ans[v]=min(ans[v],_)
    # 両替
    j=min(max_s,i+c)
    if kakutei[v][j]==0:
      dp[v][j]=min(dp[v][j],dp[v][i]+d)
      heapq.heappush(todo,[dp[v][j],v,j])
    # 鉄道
    l=g[v]
    for u,a,b in l:
      if i-a>=0 and kakutei[u][i-a]==0:
        dp[u][i-a]=min(dp[u][i-a],dp[v][i]+b)
        heapq.heappush(todo,[dp[u][i-a],u,i-a])
  print(*ans[1:],sep='\n')
if __name__=='__main__':
  main()
