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
n,m,s = map(int, input().split())
uvab=[list(map(int,input().split())) for i in range(m)]
cd=[list(map(int,input().split())) for i in range(n)]
chg=10**5

from collections import defaultdict
link = defaultdict(list)
for node in range(n):
    cmai,dmin = cd[node]

    for frm in range(2600):
        link[node*chg + frm].append([node*chg + frm+cmai, dmin])

for i in range(m):
    u,v,a,b = uvab[i]
    u-=1
    v-=1
    for sil in range(2600):
        link[u*chg + sil+a].append([v*chg + sil,b])
        link[v*chg + sil+a].append([u*chg + sil,b])

from heapq import heappush, heappop,heappushpop,heapify
from collections import defaultdict
def dks(g,start):
    INF=float('inf')
    visited = set()
    hq = []
    heappush(hq, (0,start))

    cost = defaultdict(lambda: float("inf"))
    while hq:
        shortest, i = heappop(hq)
        if i in visited:
            continue
        visited.add(i)
        cost[i] = shortest
        for j, t in g[i]:
            if j in visited:
                continue
            heappush(hq, (shortest + t, j))
    return cost

f_sil = min(s,2555)
res=dks(link,f_sil)

for node in range(1,n):
    ans=float("inf")
    for sil in range(2600):
        ans=min(ans,res[node*chg + sil])
    print(ans)
