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
import sys

def main():

    n,m,s = map(int,input().split())
    uvab = [list(map(int,input().split())) for _ in range(m)]
    cd = [list(map(int,input().split())) for _ in range(n)]

    edges = [[[] for _ in range(n)] for _ in range(2501)]

    for u,v,a,b in uvab:
        for i in range(a,2501):
            edges[i][u-1].append(((v-1, i-a), b))
            edges[i][v-1].append(((u-1, i-a), b))

    for i in range(n):
        c,d = cd[i]
        for j in range(2501):
            edges[j][i].append(((i, min(j+c, 2500)), d))

    def dijkstra(edges, c):
        hq = []
        d = [[10**20] * n for _ in range(c)]
        d[min(2500, s)][0] = 0
        heapq.heappush(hq, (0, (0, min(2500, s))))
        while hq:
            d1, p = heapq.heappop(hq)
            for p2, d2 in edges[p[1]][p[0]]:
                if d[p2[1]][p2[0]] == -1 or d[p2[1]][p2[0]] > d1 + d2:
                    d[p2[1]][p2[0]] = d1 + d2
                    heapq.heappush(hq, (d1+d2, p2))
        return d

    dis = dijkstra(edges, 2501)
    ans = [10**20]*n
    for i in range(2501):
        for j in range(n):
            ans[j] = min(ans[j], dis[i][j])

    for i in ans[1:]:
        print(i)

if __name__ == '__main__':
    main()
