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
MOD = 10 ** 9 + 7
INF = 10 ** 13
import sys
sys.setrecursionlimit(100000000)
dy = (-1,0,1,0)
dx = (0,1,0,-1)
from heapq import heapify,heappop,heappush

def dijkstra(G,s = 0): #sを始点とした最短経を求める
    n = len(G)
    dist = [INF] * n
    dist[s] = 0

    q = [(0,s)]
    heapify(q)

    while q:
        d,v = heappop(q)
        if dist[v] < d:
            continue
        for e,cost in G[v]:
            if dist[e] > dist[v] + cost:
                dist[e] = dist[v] + cost
                heappush(q,(dist[e],e))
    
    return dist

def main():
    n,m,s = map(int,input().split())
    MAXV = 2500*n
    G = [[] for _ in range(MAXV)]
    for _ in range(m):
        u,v,a,b = map(int,input().split())
        u -= 1
        v -= 1
        for i in range(2500 - a):
            G[u*2500 + i + a].append((v*2500 + i,b))
            G[v*2500 + i + a].append((u*2500 + i,b))
    
    for i in range(n):
        c,d = map(int,input().split())
        for j in range(2500 - c):
            G[i*2500 + j].append((i*2500 + j + c,d))
  
    dist = dijkstra(G,min(s,2499))
    for i in range(1,n):
        ans = min(dist[2500*i:2500*(i + 1)])
        print(ans) 

if __name__ == '__main__':
    main()
