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 os
from collections import defaultdict

import sys

if os.getenv("LOCAL"):
    sys.stdin = open("_in.txt", "r")

sys.setrecursionlimit(10 ** 9)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
# MOD = 998244353


N, M, S = list(map(int, sys.stdin.buffer.readline().split()))
UVAB = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(M)]
CD = [list(map(int, sys.stdin.buffer.readline().split())) for _ in range(N)]
CD = [(0, 0)] + CD

graph = [[] for _ in range(N + 1)]
for u, v, a, b in UVAB:
    graph[u].append((a, b, v))
    graph[v].append((a, b, u))
for vs in graph:
    vs.sort()

mada = set(list(range(2, N + 1)))
heap = [(0, 1, S)]
ans = [INF] * (N + 1)
ans[1] = 0
# dp[v, s]: v にお金を s 持ってる状態で来たときの最小値
dp = defaultdict(lambda: INF)
dp[1, S] = 0
while heap and mada:
    d, v, s = heapq.heappop(heap)
    if d > dp[v, s]:
        continue
    if v in mada:
        mada.remove(v)

    ans[v] = min(ans[v], d)
    for a, b, u in graph[v]:
        if s - a < 0:
            break
        if d + b < dp[u, s - a]:
            dp[u, s - a] = d + b
            heapq.heappush(heap, (d + b, u, s - a))
    c, dd = CD[v]
    sc = min(s + c, 50 * N + 10)
    if d + dd < dp[v, sc]:
        dp[v, sc] = d + dd
        heapq.heappush(heap, (d + dd, v, sc))

print(*ans[2:], sep='\n')
