There is a field with plants — a grid with $N$ rows (numbered $1$ through $N$) and $M$ columns (numbered $1$ through $M$); out of its $NM$ cells, $K$ cells contain plants, while the rest contain weeds. Two cells are adjacent if they have a common side.
You want to build fences in the field in such a way that the following conditions hold for each cell that contains a plant:
- it is possible to move from this cell to each adjacent cell containing a plant without crossing any fences
- it is impossible to move from this cell to any cell containing weeds or to leave the grid without crossing any fences
The fences can only be built between cells or on the boundary of the grid, i.e. on the sides of cells. The total length of the built fences is the number of pairs of side-adjacent cells such that there is a fence built on their common side plus the number of sides of cells on the boundary of the grid which have fences built on them. Find the minimum required total length of fences that need to be built.

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains three space-separated integers $N$, $M$ and $K$.
- $K$ lines follow. Each of these lines contains two space-separated integers $r$ and $c$ denoting that the cell in row $r$ and column $c$ contains a plant.

-----Output-----
For each test case, print a single line containing one integer — the minimum required length of fences.

-----Constraints-----
- $1 \le T \le 10$
- $1 \le N, M \le 10^9$
- $1 \le K \le 10^5$
- $1 \le r \le N$
- $1 \le c \le M$
- the cells containing plants are pairwise distinct

-----Subtasks-----
Subtask #1 (30 points): $1 \le N, M \le 1,000$
Subtask #2 (70 points): original constraints

-----Example Input-----
2
4 4 9
1 4
2 1 
2 2
2 3
3 1
3 3
4 1
4 2
4 3
4 4 1
1 1

-----Example Output-----
20
4

-----Explanation-----
Example case 1: The field looks like this ('x' denotes a cell containing a plant, '.' denotes a cell containing weeds):
...x
xxx.
x.x.
xxx.

An optimal solution is to build fences around the topmost plant (with length $4$), around the remaining eight plants (with length $12$) and around the hole between them (with length $4$). The total length is $4+12+4 = 20$.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'def check(x,y,mat):\n tc=0\n if(mat[(x-1)][y]==0):\n  tc+=1\n if(mat[(x)][y-1]==0):\n  tc+=1\n if(mat[(x)][y+1]==0):\n  tc+=1\n if(mat[(x+1)][y]==0):\n  tc+=1\n #print(x,y,mat[x],tc)\n return tc\n\n\n\nT=int(input())\nfor t in range(T):\n ls=list(map(int,input().split()))\n N=ls[0]\n M=ls[1]\n K=ls[2]\n mat=[[0 for i in range(M+2)] for i in range(N+2)]\n rows=[]\n cols=[] \n for k in range(K):\n  es=list(map(int,input().split()))\n  mat[es[0]][es[1]]=1\n  rows.append(es[0])\n  cols.append(es[1])\n #for row in mat:\n #   print(row)\n count=0\n for i in range(len(rows)):\n  count=count+check(rows[i],cols[i],mat)\n print(count)\n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13577