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:

    There seems to be no choice other than calculating all possible orders.
    However, there is a technique to avoid calculating the permutations.

    Use a n-bits integer as the selected items so far.
    If the i-th bit is 1, then the i-th item is chosen. Otherwise, the
    item is not chosen yet.

    Then, use this n-bits integer as the index of the array that keeps
    the previous calculations.
"""


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


def main():
    x0, y0 = map(int, input().split())
    n = int(input())
    size = 1 << n

    # coordinates of the i-th item and the time from the handbag
    xyc = []
    for o1 in range(n):
        x1, y1 = map(int, input().split())
        t1 = dist2(x1 - x0, y1 - y0)
        xyc.append((x1, y1, t1))
    # round trip time to pick two items at once
    pairs = []
    for o1 in range(n):
        x1, y1, t1 = xyc[o1]
        others = []
        for o2 in range(o1 + 1, n):
            x2, y2, t2 = xyc[o2]
            others.append((1 << o2, (t1 + dist2(x2 -x1, y2 - y1) + t2)))
        pairs.append((1 << o1, t1 * 2, others))

    # the best time of the selected items
    best = [0] * size
    # the best items to take
    pick = [0] * size

    # find the solution
    for choice in range(1, size):
        for b1, t1, others in pairs:
            if not (b1 & choice):
                continue
            rest = choice - b1
            # pick up only one item
            cost = best[rest] + t1
            items = b1
            # pick up two items at once
            for b2, t2 in others:
                if not (b2 & choice):
                    continue
                curr = best[rest - b2] + t2
                if curr < cost:
                    cost = curr
                    items = b1 + b2
            best[choice] = cost
            pick[choice] = items
            break

    def order():
        yield 0
        p = size - 1
        while p:
            m = pick[p]
            p -= m
            k = 1
            while m:
                if m & 1:
                    yield k
                m >>= 1
                k += 1
            yield 0

    print(best[-1])
    print(" ".join(map(str, order())))


if __name__ == '__main__':
    main()
