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
INF = float('inf')

N, M, S = map(int,input().split())
graph = [[] for _ in range(N)]
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))

C, D = [0]*N, [0]*N
for i in range(N):
    c, d = map(int,input().split())
    C[i], D[i] = c, d

def Extdijkstra(start):
    global S
    N_d = len(graph)
    M = 2500

    if S >= M:
        S = M-1
    
    dist = [[INF]*M for _ in range(N_d)]
    dist[start][S] = 0
    q = [(0, S, start)]

    while q:
        d, num, v = heappop(q)
        if num+C[v] < M and d+D[v] < dist[v][num+C[v]]:  # 自己ループ
            dist[v][num+C[v]] = d+D[v]
            heappush(q, (d+D[v], num+C[v], v))
        for node, c, t in graph[v]:
            if num >= c and d + t < dist[node][num-c]:
                dist[node][num-c] = d+t
                heappush(q, (d+t, num-c, node))
    return dist

d = Extdijkstra(0)
for i in range(1, N):
	print(min(d[i]))