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 heapq
def dijkstra(s, n, g): # s: start, n: |V|, g; glaph 
    INF = 10**18
    d = [INF] * n
    d[s] = 0
    que = [] # (a,b): a... shortest dist, b... v
    heapq.heappush(que, (0, s))

    while que:
        dist, v = heapq.heappop(que)
        if d[v] < dist: continue # if v has been already used -> continue
        for next_v, cost in g[v]:
            if d[next_v] > d[v] + cost:
                d[next_v] = d[v] + cost
                heapq.heappush(que, (d[next_v], next_v))
    return d


def main():
    n,m,s = map(int, input().split())
    g = [ [] for _ in range(n*2501+1)]
    for _ in range(m):
        u,v,a,b = map(int, input().split())
        u-=1
        v-=1
        for j in range(2501):
            if j < a: continue
            g[u*2501+j].append((v*2501+j-a,b))
            g[v*2501+j].append((u*2501+j-a,b))


    cdl = [(-1,-1)]
    for i in range(n):
        c,d = map(int, input().split())
        for j in range(2501):
            maxc = min(2500,j+c)
            g[i*2501+j].append((i*2501+maxc,d))

    s = min(2500,s)
    dist = dijkstra(s, n*2501+1, g)
    # print(dist)
    for i in range(1,n):
        ans = min(dist[i*2501:i*2501+2500])
        print(ans)


if __name__ == "__main__":
    main()