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
# ABC 164 E 草案
import sys
sys.setrecursionlimit(10**6)

n,m,s= map(int, input().split())
g= [list(map(int, input().split())) for i in range(m)]
# 両替所
c= [list(map(int, input().split())) for i in range(n)]

x=2501
# 作るグラフ (辺*所有銀貨,必要時間)
gg=[[] for i in range(n*x)]
for i in range(n):
    for j in range(x-1):
        # 金貨を銀貨に交換
        y=j+c[i][0]
        if y>=x-1:
            y=x-1
        gg[i*x+j].append((i*x+y,c[i][1]))


for i in range(1,x):
    for u,v,a,b in g:
        u-=1
        v-=1
        # 銀貨足りる時移動できる
        if i-a>=0:
            gg[v*x+i].append((u*x+i-a,b))
            gg[u*x+i].append((v*x+i-a,b))




#ダイクストラ法
from heapq import heappush, heappop

def dijkstra(edges, size, source):
    # edge=探索するグラフ
    # size=グラフの点の数
    # source=スタート点
    distance = [float('inf')] * size
    distance[source] = 0
    visited = [False] * size
    pq = []
    heappush(pq, (0, source))
    while pq:
        dist_v, v = heappop(pq)
        visited[v] = True
        for u, weight in edges[v]:
            if not visited[u]:
                new_dist = dist_v + weight
                if distance[u] > new_dist:
                    distance[u] = new_dist
                    heappush(pq, (new_dist, u))
    return distance
s=min(s,2500)
ans=dijkstra(gg,n*x,s)

for i in range(1, n):
    print(min(ans[x * i:x * (i + 1)]))