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
import sys
read = sys.stdin.read
readline = sys.stdin.readline
readlines = sys.stdin.readlines

# グラフを拡張してダイクストラ法を使う

def main():
    N,M,S = map(int, readline().split())

    E = [[] for _ in range(N+1)]
    for _ in range(M):
        u,v,a,b = map(int, readline().split())
        E[u].append((v, a, b))
        E[v].append((u, a, b))

    cd = map(int, read().split())
    CD = []
    for c, d in zip(*[iter(cd)]*2):
        CD.append((c, d))

    # Dijkstra
    # d[v][s] := 頂点vで,所持金sとなる最小の所要時間
    if S > 2500:
        S = 2500 - 1
    INF = 10**15
    d = [[INF] * 2500 for _ in range(N+1)]
    d[1][S] = 0

    Q = []
    heappush(Q, (0, S, 1))  # (所要時間, 所持金, 頂点)
    
    while Q:
        time, s, u = heappop(Q)
        if time > d[u][s]:
            continue
        
        c_u, d_u = CD[u-1]
        if s + c_u < 2500 and time + d_u < d[u][s + c_u]:
            d[u][s + c_u] = time + d_u
            heappush(Q, (time + d_u, s + c_u, u))
        
        for v, a, b in E[u]:
            if s >= a and time + b < d[v][s - a]:
                d[v][s - a] = time + b
                heappush(Q, (time + b, s - a, v))

    for i in range(2,N+1):
        print(min(d[i]))


if __name__ == "__main__":
    main()
