The capital of Berland looks like a rectangle of size n × m of the square blocks of same size.

Fire!

It is known that k + 1 blocks got caught on fire (k + 1 ≤ n·m). Those blocks are centers of ignition. Moreover positions of k of these centers are known and one of these stays unknown. All k + 1 positions are distinct.

The fire goes the following way: during the zero minute of fire only these k + 1 centers of ignition are burning. Every next minute the fire goes to all neighbouring blocks to the one which is burning. You can consider blocks to burn for so long that this time exceeds the time taken in the problem. The neighbouring blocks are those that touch the current block by a side or by a corner.

Berland Fire Deparment wants to estimate the minimal time it takes the fire to lighten up the whole city. Remember that the positions of k blocks (centers of ignition) are known and (k + 1)-th can be positioned in any other block.

Help Berland Fire Department to estimate the minimal time it takes the fire to lighten up the whole city.

Input

The first line contains three integers n, m and k (1 ≤ n, m ≤ 109, 1 ≤ k ≤ 500).

Each of the next k lines contain two integers xi and yi (1 ≤ xi ≤ n, 1 ≤ yi ≤ m) — coordinates of the i-th center of ignition. It is guaranteed that the locations of all centers of ignition are distinct.

Output

Print the minimal time it takes the fire to lighten up the whole city (in minutes).

Examples

Input

7 7 3
1 2
2 1
5 5


Output

3


Input

10 5 1
3 3


Output

2

Note

In the first example the last block can have coordinates (4, 4).

In the second example the last block can have coordinates (8, 3).
import sys
from collections import Counter
from operator import itemgetter
from heapq import heappop, heappush

n, m, k = map(int, input().split())
points = [list(map(int, line.split())) for line in sys.stdin]
pts_sorted_x = sorted(points)
pts_sorted_y = sorted(points, key=itemgetter(1, 0))
inf = 10**9+1
OK = (inf, inf)


def solve2(imos, t):
    acc, cur = 0, 0

    for k in sorted(imos.keys()):
        if t < k:
            break
        if acc <= 0 and cur+1 < k or acc + imos[k] <= 0:
            acc = 0
            break
        acc += imos[k]

    return acc <= 0


def add_imos(imos, x, y):
    imos[x] += y
    if imos[x] == 0:
        del imos[x]


def solve(t, px=-1, py=-1):
    set_x = {1, n}
    set_y = {1, m}

    for x, y in points:
        set_x.update((max(1, x-t), max(1, x-t-1), min(n, x+t), min(n, x+t+1)))
        set_y.update((max(1, y-t), max(1, y-t-1), min(m, y+t), min(m, y+t+1)))

    ans_x = ans_y = inf
    pi, imos, hq = 0, Counter(), []
    if px != -1:
        imos[py] += 1
        imos[py+t*2+1] -= 1

    for cx in sorted(set_x):
        while hq and hq[0][0] < cx:
            add_imos(imos, hq[0][1], -1)
            add_imos(imos, hq[0][2], +1)
            heappop(hq)
        while pi < k and pts_sorted_x[pi][0]-t <= cx <= pts_sorted_x[pi][0]+t:
            x, y = pts_sorted_x[pi]
            add_imos(imos, max(1, y-t), 1)
            add_imos(imos, y+t+1, -1)
            heappush(hq, (x+t, max(1, y-t), y+t+1))
            pi += 1

        if solve2(imos, m):
            ans_x = cx
            break

    pi = 0
    imos.clear()
    hq.clear()
    if px != -1:
        imos[px] += 1
        imos[px+t*2+1] -= 1

    for cy in sorted(set_y):
        while hq and hq[0][0] < cy:
            add_imos(imos, hq[0][1], -1)
            add_imos(imos, hq[0][2], +1)
            heappop(hq)
        while pi < k and pts_sorted_y[pi][1]-t <= cy <= pts_sorted_y[pi][1]+t:
            x, y = pts_sorted_y[pi]
            add_imos(imos, max(1, x-t), 1)
            add_imos(imos, x+t+1, -1)
            heappush(hq, (y+t, max(1, x-t), x+t+1))
            pi += 1

        if solve2(imos, n):
            ans_y = cy
            break

    return ans_x, ans_y


ok, ng = 10**9+1, -1
while abs(ok - ng) > 1:
    mid = (ok + ng) >> 1
    p = solve(mid)
    if p == OK:
        ok = mid
        continue

    if solve(mid, p[0], p[1]) == OK:
        ok = mid
    else:
        ng = mid

print(ok)
