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
import heapq
def dijkstra(s,es):
    INF=sys.maxsize
    cs=[INF]*len(es)
    cs[s],hq=0,[]
    heapq.heapify(hq)
    heapq.heappush(hq,(cs[s],s))
    while hq:
        d,v=heapq.heappop(hq)
        if cs[v]<d:continue
        for t,b in es[v]:
            if cs[t]>cs[v]+b:
                cs[t]=cs[v]+b
                heapq.heappush(hq,(cs[t],t))
    return cs
def nwc(n,c):
    MAX_COINS=2500
    return (MAX_COINS+1)*n+c
def main():
    MAX_COINS=2500
    N,M,S=tuple(map(int,sys.stdin.readline().split()))
    S=S if S<MAX_COINS else MAX_COINS
    es=[[] for _ in range(N*(MAX_COINS+1))]
    for _ in range(M):
        u,v,a,b=tuple(map(int,sys.stdin.readline().split()))
        u,v=u-1,v-1
        for c in range(a,MAX_COINS+1):
            es[nwc(u,c)].append((nwc(v,c-a),b))
            es[nwc(v,c)].append((nwc(u,c-a),b))
    for n in range(N):
        c,d=tuple(map(int,sys.stdin.readline().split()))
        c=c if c<MAX_COINS else MAX_COINS
        for i in range(0,MAX_COINS-c+1):es[nwc(n,i)].append((nwc(n,i+c),d))
    cs=dijkstra(nwc(0,S),es)
    for n in range(1,N):print(min(cs[n*(MAX_COINS+1):(n+1)*(MAX_COINS+1)]))
if __name__=='__main__':main()