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 sys
def input(): return sys.stdin.readline().strip()
def mapint(): return map(int, input().split())
sys.setrecursionlimit(10**9)

N, M, S = mapint()
S = min(S, N*50)
query = [[[] for _ in range(N*50+1)] for _ in range(N)]
for _ in range(M):
    u, v, a, b = mapint()
    for coin in range(a, N*50+1):
        query[u-1][coin].append((v-1, a, b))
        query[v-1][coin].append((u-1, a, b))
for i in range(N):
    c, d = mapint()
    for coin in range(N*50+1):
        query[i][coin].append((i, -c, d))

inf = 10**18
dp = [[inf]*(N*50+1) for _ in range(N)]
dp[0][0] = 0

import heapq
queue = [(0, 0, S)]
checked = [0]*N
cnt = 0

while queue:
    t, now, coin = heapq.heappop(queue)
    if cnt==N:
        break
    if not checked[now]:
        checked[now] = 1
        cnt += 1
    qlis = query[now][coin]
    for nx, a, b in qlis:
        if 0<=coin-a:
            coin = min(coin, N*50+a)
            if t+b<dp[nx][coin-a]:
                dp[nx][coin-a] = t+b
                heapq.heappush(queue, (t+b, nx, coin-a))

for i in range(1, N):
    print(min(dp[i]))
