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
readline = sys.stdin.readline
N, M, S = map(int, readline().split())

from heapq import heappop as hpp, heappush as hp
def dijkstra(N, s, Edge):
    # Edge は　(cost, vf) の形
    inf = 10**18
    dist = [inf] * N
    dist[s] = 0
    Q = [(0, s)]
    while Q:
        dn, vn = hpp(Q)
        if dn > dist[vn]:
            continue
        for df, vf in Edge[vn]:
            if dist[vn] + df < dist[vf]:
                dist[vf] = dist[vn] + df
                hp(Q, (dn + df,vf))
    return dist

geta = 2451
Edge = [[] for _ in range(N*geta)]
S = min(S, geta - 1)
for m in range(M):
    u, v, a, b = map(int, readline().split())
    u -= 1
    v -= 1
    for mon in range(a, geta):
        Edge[u*geta+mon].append((b, v*geta + (mon - a)))
        Edge[v*geta+mon].append((b, u*geta + (mon - a)))

for n in range(N):
    c, d = map(int, readline().split())
    for mon in range(geta):
        Edge[n*geta + mon].append((d, n*geta + min(geta-1, mon + c)))

dist = dijkstra(N*geta, S, Edge) 
INF = 10**18
ans = [INF]*N
for i in range(N*geta):
    ans[i//geta] = min(ans[i//geta], dist[i])
print('\n'.join(map(str, ans[1:])))