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
from heapq import heappop, heappush

sys.setrecursionlimit(10 ** 7)
rl = sys.stdin.readline


def solve():
    N, M, S = map(int, rl().split())
    init_s = min(2500, S)
    graph = [[] for _ in range(N)]
    for _ in range(M):
        u, v, a, b = map(int, rl().split())
        u, v = u - 1, v - 1
        graph[u].append((v, a, b))
        graph[v].append((u, a, b))
    C, D = [0] * N, [0] * N
    for i in range(N):
        C[i], D[i] = map(int, rl().split())
    
    costs = [[10 ** 18] * 2501 for _ in range(N)]
    costs[0][init_s] = 0
    
    que = [(0, 0, init_s)]
    while que:
        t, cur, s = heappop(que)
        self_roop_s = min(s + C[cur], 2500)
        self_roop_t = t + D[cur]
        if self_roop_t < costs[cur][self_roop_s]:
            costs[cur][self_roop_s] = self_roop_t
            heappush(que, (self_roop_t, cur, self_roop_s))
        for child, s_cost, t_cost in graph[cur]:
            if s < s_cost:
                continue
            nt = t + t_cost
            ns = s - s_cost
            if nt < costs[child][ns]:
                costs[child][ns] = nt
                heappush(que, (nt, child, ns))
    
    for costs_i in costs[1:]:
        print(min(costs_i))


if __name__ == '__main__':
    solve()
