Girl Lena likes it when everything is in order, and looks for order everywhere. Once she was getting ready for the University and noticed that the room was in a mess — all the objects from her handbag were thrown about the room. Of course, she wanted to put them back into her handbag. The problem is that the girl cannot carry more than two objects at a time, and cannot move the handbag. Also, if he has taken an object, she cannot put it anywhere except her handbag — her inherent sense of order does not let her do so.

You are given the coordinates of the handbag and the coordinates of the objects in some Сartesian coordinate system. It is known that the girl covers the distance between any two objects in the time equal to the squared length of the segment between the points of the objects. It is also known that initially the coordinates of the girl and the handbag are the same. You are asked to find such an order of actions, that the girl can put all the objects back into her handbag in a minimum time period.
The first line of the input file contains the handbag's coordinates *x*<sub class="lower-index">*s*</sub>, *y*<sub class="lower-index">*s*</sub>. The second line contains number *n* (1 ≤ *n* ≤ 24) — the amount of objects the girl has. The following *n* lines contain the objects' coordinates. All the coordinates do not exceed 100 in absolute value. All the given positions are different. All the numbers are integer.
none
0 0
2
1 1
-1 1
8
0 1 2 0 
1 1
3
4 3
3 4
0 0
32
0 1 2 0 3 0 
-3 4
1
2 2
58
0 1 0 
7 -7
2
3 1
-3 8
490
0 1 2 0 
3 -9
3
0 -9
-10 -3
-12 -2
502
0 1 0 2 3 0 
4 -1
4
14 -3
-11 10
-3 -5
-8 1
922
0 1 0 2 4 0 3 0 
7 -11
5
-1 7
-7 -11
12 -4
8 -6
-18 -8
1764
0 1 3 0 2 5 0 4 0 
11 3
6
-17 -17
-4 -9
15 19
7 4
13 1
5 -6
2584
0 1 2 0 3 0 4 6 0 5 0 
-6 4
7
-10 -11
-11 -3
13 27
12 -22
19 -17
21 -21
-5 4
6178
0 1 4 0 2 0 3 7 0 5 6 0 
27 -5
8
-13 -19
-20 -8
11 2
-23 21
-28 1
11 -12
6 29
22 -15
14062
0 1 2 0 3 7 0 4 5 0 6 8 0 
31 9
9
8 -26
26 4
3 2
24 21
14 34
-3 26
35 -25
5 20
-1 8
9384
0 1 7 0 2 0 3 9 0 4 5 0 6 8 0 
-44 47
24
96 -18
-50 86
84 68
-25 80
-11 -15
-62 0
-42 50
-57 11
-5 27
-44 67
-77 -3
-27 -46
32 63
86 13
-21 -51
-25 -62
-14 -2
-21 86
-92 -94
-44 -34
-74 55
91 -35
-10 55
-34 16
191534
0 1 22 0 2 10 0 3 14 0 4 18 0 5 20 0 6 11 0 7 0 8 24 0 9 17 0 12 15 0 13 23 0 16 19 0 21 0 
5 4
11
-26 2
20 35
-41 39
31 -15
-2 -44
16 -28
17 -6
0 7
-29 -35
-17 12
42 29
19400
0 1 3 0 2 11 0 4 6 0 5 9 0 7 0 8 10 0 
-44 22
12
-28 24
41 -19
-39 -36
12 -18
-31 -24
-7 29
45 0
12 -2
42 31
28 -37
-34 -38
6 24
59712
0 1 5 0 2 10 0 3 11 0 4 8 0 6 12 0 7 9 0 
40 -36
13
3 -31
28 -43
45 11
47 -37
47 -28
-30 24
-46 -33
-31 46
-2 -38
-43 -4
39 11
45 -1
50 38
52988
0 1 9 0 2 0 3 13 0 4 5 0 6 8 0 7 10 0 11 12 0 
-54 2
14
-21 -2
-5 34
48 -55
-32 -23
22 -10
-33 54
-16 32
-53 -17
10 31
-47 21
-52 49
34 42
-42 -25
-32 31
55146
0 1 4 0 2 7 0 3 5 0 6 11 0 8 13 0 9 12 0 10 14 0 
-19 -31
15
-31 -59
60 -34
-22 -59
5 44
26 39
-39 -23
-60 -7
1 2
-5 -19
-41 -26
46 -8
51 -2
60 4
-12 44
14 49
60546
0 1 3 0 2 11 0 4 14 0 5 15 0 6 0 7 10 0 8 9 0 12 13 0 
-34 19
16
-44 24
30 -42
46 5
13 -32
40 53
35 49
-30 7
-60 -50
37 46
-18 -57
37 -44
-61 58
13 -55
28 22
-50 -3
5 52
81108
0 1 12 0 2 11 0 3 14 0 4 13 0 5 9 0 6 16 0 7 15 0 8 10 0 
-64 -6
17
-3 -18
66 -58
55 34
-4 -40
-1 -50
13 -9
56 55
3 42
-54 -52
51 -56
21 -27
62 -17
54 -5
-28 -24
12 68
43 -22
8 -6
171198
0 1 14 0 2 10 0 3 7 0 4 5 0 6 17 0 8 15 0 9 0 11 16 0 12 13 0 
7 -35
18
24 -3
25 -42
-56 0
63 -30
18 -63
-30 -20
-53 -47
-11 -17
-22 -54
7 -41
-32 -3
-29 15
-30 -25
68 15
-18 70
-28 19
-12 69
44 29
70504
0 1 4 0 2 5 0 3 11 0 6 13 0 7 9 0 8 0 10 0 12 16 0 14 18 0 15 17 0 
-8 47
19
47 51
43 -57
-76 -26
-23 51
19 74
-36 65
50 4
48 8
14 -67
23 44
5 59
7 -45
-52 -6
-2 -33
34 -72
-51 -47
-42 4
-41 55
22 9
112710
0 1 10 0 2 15 0 3 16 0 4 0 5 11 0 6 18 0 7 8 0 9 12 0 13 17 0 14 19 0 
44 75
20
-19 -33
-25 -42
-30 -61
-21 44
7 4
-38 -78
-14 9
65 40
-27 25
65 -1
-71 -38
-52 57
-41 -50
-52 40
40 44
-19 51
42 -43
-79 -69
26 -69
-56 44
288596
0 1 19 0 2 13 0 3 6 0 4 16 0 5 7 0 8 15 0 9 14 0 10 17 0 11 18 0 12 20 0 
42 -34
21
4 62
43 73
29 -26
68 83
0 52
-72 34
-48 44
64 41
83 -12
-25 52
42 59
1 38
12 -79
-56 -62
-8 67
84 -83
22 -63
-11 -56
71 44
7 55
-62 65
196482
0 1 15 0 2 4 0 3 0 5 12 0 6 21 0 7 10 0 8 19 0 9 16 0 11 20 0 13 17 0 14 18 0 
-44 42
22
-67 -15
74 -14
67 76
-57 58
-64 78
29 33
-27 27
-20 -52
-54 -2
-29 22
31 -65
-76 -76
-29 -51
-5 -79
-55 36
72 36
-80 -26
5 60
-26 69
78 42
-47 -84
8 83
181122
0 1 17 0 2 16 0 3 20 0 4 5 0 6 18 0 7 10 0 8 13 0 9 15 0 11 14 0 12 21 0 19 22 0 
52 92
23
-67 -82
31 82
-31 -14
-1 35
-31 -49
-75 -14
78 -51
-35 -24
28 -84
44 -51
-37 -9
-38 -91
41 57
-19 35
14 -88
-60 -60
-13 -91
65 -8
-30 -46
72 -44
74 -5
-79 31
-3 84
492344
0 1 12 0 2 23 0 3 11 0 4 14 0 5 16 0 6 22 0 7 20 0 8 19 0 9 10 0 13 0 15 17 0 18 21 0 
-21 -47
24
-37 1
-65 8
-74 74
58 -7
81 -31
-77 90
-51 10
-42 -37
-14 -17
-26 -71
62 45
56 43
-75 -73
-33 68
39 10
-65 -93
61 -93
30 69
-28 -53
5 24
93 38
-45 -14
3 -86
63 -80
204138
0 1 22 0 2 7 0 3 6 0 4 5 0 8 19 0 9 20 0 10 23 0 11 21 0 12 15 0 13 16 0 14 18 0 17 24 0 
31 16
21
-9 24
-59 9
-25 51
62 52
39 15
83 -24
45 -81
42 -62
57 -56
-7 -3
54 47
-14 -54
-14 -34
-19 -60
-38 58
68 -63
-1 -49
6 75
-27 22
-58 -77
-10 56
121890
0 1 10 0 2 19 0 3 15 0 4 11 0 5 0 6 16 0 7 9 0 8 17 0 12 13 0 14 20 0 18 21 0 
20 -1
22
-51 -31
-41 24
-19 46
70 -54
60 5
-41 35
73 -6
-31 0
-29 23
85 9
-7 -86
8 65
-86 66
-35 14
11 19
-66 -34
-36 61
84 -10
-58 -74
-11 -67
79 74
3 -67
135950
0 1 16 0 2 6 0 3 12 0 4 18 0 5 7 0 8 14 0 9 15 0 10 21 0 11 22 0 13 17 0 19 20 0 
-49 4
23
-18 -53
-42 31
18 -84
-20 -70
-12 74
-72 81
12 26
3 9
-70 -27
34 -32
74 -47
-19 -35
-46 -8
-77 90
7 -42
81 25
84 81
-53 -49
20 81
-39 0
-70 -44
-63 77
-67 -73
169524
0 1 4 0 2 22 0 3 15 0 5 19 0 6 14 0 7 8 0 9 21 0 10 11 0 12 13 0 16 17 0 18 23 0 20 0 
-81 35
24
58 27
92 -93
-82 63
-55 80
20 67
33 93
-29 46
-71 -51
-19 8
58 -71
13 60
0 -48
-2 -68
-56 53
62 52
64 32
-12 -63
-82 -22
9 -43
55 12
77 -21
26 -25
-91 -32
-66 57
337256
0 1 16 0 2 10 0 3 24 0 4 14 0 5 11 0 6 15 0 7 9 0 8 17 0 12 13 0 18 23 0 19 22 0 20 21 0 
45 79
24
-66 22
10 77
74 88
59 1
-51 -86
-60 91
1 -51
-23 85
3 96
38 -4
-55 43
9 -68
-4 83
75 -13
64 -74
28 27
92 -57
-20 -64
30 -44
-95 67
13 55
67 -4
42 77
61 87
277576
0 1 11 0 2 9 0 3 24 0 4 16 0 5 18 0 6 20 0 7 12 0 8 13 0 10 19 0 14 22 0 15 17 0 21 23 0 
-61 34
24
-57 -46
-37 -24
-87 -54
51 -89
-90 2
95 -63
-24 -84
-85 38
-52 -62
96 4
89 -22
-16 -3
-2 -14
71 -62
-51 68
-83 -24
15 77
-61 45
17 -32
-68 -87
-93 -28
-85 24
-84 -34
-4 1
262400
0 1 9 0 2 12 0 3 23 0 4 19 0 5 22 0 6 14 0 7 20 0 8 18 0 10 11 0 13 24 0 15 17 0 16 21 0 
70 90
24
-64 -96
-87 -82
10 -65
94 22
95 60
-13 54
-83 -92
95 -50
-65 -91
96 -88
80 -56
-31 85
58 86
-28 22
-22 45
-24 -12
-62 70
-2 -77
-31 -72
61 37
67 43
-5 -30
-84 -59
-91 51
585696
0 1 7 0 2 23 0 3 18 0 4 8 0 5 0 6 12 0 9 19 0 10 11 0 13 0 14 15 0 16 22 0 17 24 0 20 21 0 
72 -37
24
56 -47
-37 -20
76 -46
-14 11
-63 -46
52 74
-60 -23
27 8
-78 -26
15 -23
74 -90
39 -64
86 53
77 11
-47 -44
-1 -14
90 56
76 -88
-27 51
-67 -8
-27 4
83 -91
54 68
56 26
224008
0 1 3 0 2 7 0 4 19 0 5 15 0 6 23 0 8 10 0 9 20 0 11 12 0 13 17 0 14 24 0 16 21 0 18 22 0 
9 -5
10
-22 23
22 -26
10 -32
18 -34
7 -27
2 -38
-5 -24
-38 -15
21 -32
-17 37
13454
0 1 10 0 2 9 0 3 4 0 5 6 0 7 8 0 
#!/usr/bin/env python

"""
CodeForces: 8C. Looking for Order

Strategy

    The optimum way to pick up the pair of objects is:

        dd = distance between two objects
        d1 = distance between the handbag and one of object
        d2 = distance between the handbag and other object

        if dd < d1 + d2:
            pick up the two objects at once
            optimum time = d1 + dd + d2
        else:
            pick up the two objects one by one
            optimum time = (d1 + d2) * 2

    The optimum pair can be calculated by:

        optimum pair = min(min(dd, d1 + d2) for all possible pairs)
"""


def dist2(dx, dy):
    return (dx * dx) + (dy * dy)


def main():
    x0, y0 = map(int, input().split())
    n = int(input())

    # pre-calculate the optimum way for all possible pairs
    xyd = []
    pair = []
    for o1 in range(n):
        x1, y1 = map(int, input().split())
        d1 = dist2(x1 - x0, y1 - y0)
        xyd.append((x1, y1, d1))
        other = []
        for o2 in range(o1):
            x2, y2, d2 = xyd[o2]
            d0 = d1 + d2
            dd = dist2(x2 -x1, y2 - y1)
            other.append((min(dd, d0), o2, d0))
        other.sort(reverse=True)
        pair.append(other)

    # calculate the optimum order by finding the optimum (n/2) pairs
    order = [0]
    total = 0
    for _ in range(n // 2):
        best_d = 1000000
        best_o = 0
        for o1 in range(1, n):
            if xyd[o1] is None:
                continue
            other = pair[o1]
            while other:
                dd, o2, _ = other[-1]
                if xyd[o2] is not None:
                    if dd < best_d:
                        best_o = o1
                        best_d = dd
                    break
                other.pop()
        _, o2, d0 = pair[best_o].pop()
        total += best_d + d0
        order.append(o2 + 1)
        if best_d == d0:
            order.append(0)
        order.append(best_o + 1)
        order.append(0)
        xyd[best_o] = xyd[o2] = None

    # pick up the last object
    if n & 1:
        for o1 in range(n):
            if xyd[o1] is not None:
                total += xyd[o1][2] * 2
                order.append(o1 + 1)
                order.append(0)
                break

    print(total)
    print(" ".join(map(str, order)))


if __name__ == '__main__':
    main()
