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 heappop,heappush
import sys
input=sys.stdin.buffer.readline
N,M,S=map(int,input().split())
INF=10**18
Time=[[INF]*2501 for _ in range(N)]#[都市][お金]
graph=[[] for _ in range(N)]
S=min(S,2500)
for _ in range(M):
    u,v,a,b=map(int,input().split())
    graph[u-1].append((v-1,a,b))
    graph[v-1].append((u-1,a,b))

cd=[tuple(map(int,input().split())) for _ in range(N)]

hq=[(0,0,S)] #時間、都市、金
while hq:
    ct,cur_city,si=heappop(hq)
    for nxt_city,cost,t in graph[cur_city]:#都市、費用、所要時間
        if si>=cost:
            new_t=ct+t
            if new_t<Time[nxt_city][si-cost]:
              Time[nxt_city][si-cost]=new_t
              heappush(hq,(new_t,nxt_city,si-cost))
    c,d=cd[cur_city]
    v_new=min(si+c,2500)
    t_new=ct+d
    if t_new<Time[cur_city][v_new]:
      Time[cur_city][v_new]=t_new
      heappush(hq,(t_new,cur_city,v_new))
for i in range(1,N):
    print(min(Time[i]))