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
from heapq import heappop, heappush
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline


phi = lambda city, silver: city * 2501 + silver
psi = lambda x: (x//2501, x % 2501)


n, m, s = map(int, input().split())
g = [[] for _ in range(n*2501+1)]

for i in range(m):
    u, v, a, b = map(int, input().split())
    for j in range(2501):
        if j-a >= 0:
            g[phi(u-1, j)].append((phi(v-1, j-a), b))
            g[phi(v-1, j)].append((phi(u-1, j-a), b))

for i in range(n):
    c, d = map(int, input().split())
    for j in range(2501):
        if j + c <= 2500:
            g[phi(i, j)].append((phi(i, j+c), d))


def dijkstra(s, n, g):
    # g[i][j] : the cost of i→j  or g[i] = [(target, cost), ...]
    # must import heapq
    # sys.stdin.readline is recommended
    d = [10**18] * n
    d[s] = 0
    q = [(0, s)]
    while q:
        dist_u, u = heappop(q)
        if d[u] < dist_u:
            continue
        for v, c in g[u]:  # 'in g[u]' if g is an adj list.
            if d[v] > dist_u + c:
                d[v] = dist_u + c
                heappush(q, (d[v], v))
    return d


s = min(s, 2500)
d = dijkstra(phi(0, s), n*2501+1, g)

for i in range(1, n):
    print(min(d[phi(i, 0):phi(i+1, 0)]))
    