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
input = sys.stdin.readline

INF = 10 ** 18

class Edge():
    def __init__(self, end, cost, time):
        self.end = end
        self.cost = cost
        self.time = time

dp = []
h = []

def push(t, v, x):
    if x < 0:
        return
    if dp[v][x] <= t:
        return
    dp[v][x] = t
    heapq.heappush(h, (t, v, x))

def dijkstra(n, G, cd, start, s, max_s):
    global dp
    dp = [[INF for j in range(max_s+1)] for i in range(n)] 
    push(0, start, s)

    while h:
        # 使っていない頂点のうち、現時点で最も距離の近いものを選びvとする
        t, v, x = heapq.heappop(h)
        if dp[v][x] < t:
            continue
        # 両替
        if x < max_s:
            push(t+cd[v][1], v, min(x+cd[v][0], max_s))

        # vから到達可能な頂点について、距離が短くなれば更新
        for e in G[v]:
            push(t+e.time, e.end, x-e.cost)


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)]

es = [[] for i in range(n)]
max_s = 50 * (n-1)

for u, v, a, b in uvab:
    es[u-1].append(Edge(v-1,a,b))
    es[v-1].append(Edge(u-1,a,b))

dijkstra(n, es, cd, 0, min(max_s,s), max_s)

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


    