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
range = xrange
input = raw_input

M = 2501
inf = 3 * 10**18

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
n = inp[ii]; ii += 1
m = inp[ii]; ii += 1
s = inp[ii]; ii += 1

s = min(M - 1, s)

coupl = [[] for _ in range(n)]
V = []
A = []
B = []
for _ in range(m):
    u = inp[ii] - 1; ii += 1
    v = inp[ii] - 1; ii += 1
    coupl[u].append(2 * _)
    coupl[v].append(2 * _ + 1)
    V.append(v)
    V.append(u)

    A.append(inp[ii]); ii += 1
    B.append(inp[ii]); ii += 1

C = inp[ii + 0: ii + 2 * n: 2]
D = inp[ii + 1: ii + 2 * n: 2]
ii += n

import heapq
DP = [inf] * (n * M)
found = [0] * (n * M)
DP[s] = 0

#heapq.cmp_lt = 
queue = [(0,s)]
while queue:
    _,ind = heapq.heappop(queue)
    if found[ind]:
        continue
    found[ind] = 1
    node,money = divmod(ind, M)
    for eind in coupl[node]:
        v = V[eind]
        ind2 = ind - A[eind >> 1] + M*(v - node)
        if money >= A[eind >> 1] and not found[ind2]:
            val = DP[ind] + B[eind >> 1]
            if DP[ind2] > val:
                DP[ind2] = val
                heapq.heappush(queue, (val, ind2))

    ind2 = ind - money + min(M - 1, C[node] + money)
    val = DP[ind] + D[node]
    if DP[ind2] > val:
        DP[ind2] = val
        heapq.heappush(queue, (val, ind2))

ans = []
for node in range(1, n):
    ans.append(min(DP[node * M + m] for m in range(M)))
print '\n'.join(str(x) for x in ans)