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 heappush, heappop
def main():
    N, M, S = map(int, input().split())
    G = [[] for _ in range(N)]
    for _ in range(M):
        u, v, a, b = map(int, input().split())
        u -= 1
        v -= 1
        G[u].append((v, a, b))
        G[v].append((u, a, b))
    CD = [list(map(int, input().split())) for _ in range(N)]
    distance = [[1<<62]*2501 for _ in range(N)]
    q = [(0, 0, min(S, 2500))]
    distance[0][min(S, 2500)] = 0
    while q:
        d, v, vs = heappop(q)
        if distance[v][vs] != d:
            continue

        vc, vd = CD[v]
        ud = d + vd
        us = vs + vc
        if us > 2500:
            us = 2500
        if distance[v][us] > ud:
            distance[v][us] = ud
            heappush(q, (ud, v, us))

        for u, fee, b in G[v]:
            if fee > vs:
                continue
            ud = d + b
            us = vs - fee
            if us > 2500:
                us = 2500
            if distance[u][us] > ud:
                distance[u][us] = ud
                heappush(q, (ud, u, us))
    print("\n".join(str(min(d)) for d in distance[1:]))

main()
