Spring is interesting season of year. Chef is thinking about different things, but last time he thinks about interesting game - "Strange Matrix". 
Chef has a matrix that consists of n rows, each contains m elements. Initially, the element aij of matrix equals j. (1 ≤ i ≤ n, 1 ≤ j ≤ m). 
Then p times some element aij is increased by 1. 
Then Chef needs to calculate the following: 

- For each row he tries to move from the last element (with number m) to the first one (with the number 1). 
- While staying in aij Chef can only move to aij - 1 only if aij - 1 ≤ aij. 
- The cost of such a movement is aij - aij - 1.
- Otherwise Chef can't move and lose (in this row).
- If Chef can move from the last element of the row to the first one, then the answer is the total cost of all the movements. 
- If Chef can't move from the last element of the row to the first one, then the answer is -1. 

Help Chef to find answers for all the rows after P commands of increasing. 

-----Input-----

- The first line contains three integers n, m and p denoting the number of rows, the number of elements a single row and the number of increasing commands. 
- Each of next p lines contains two integers i and j denoting that the element aij  is increased by one. 

-----Output-----
- For each row in a single line print the answer after the P increasing commands.

-----Constraints-----
- 1 ≤ n, m, p ≤ 10 ^ 5
- 1 ≤ i ≤ n
- 1 ≤ j ≤ m

-----Example-----
Input:
4 4 6
2 2
3 2 
3 2 
4 3
4 4
4 3

Output:
3
3
-1
4


-----Explanation-----

Here is the whole matrix after P commands:
1 2 3 4
1 3 3 4
1 4 3 4
1 2 5 5

Explanations to the answer: 
- The first line is without changes: 4-3=1, 3-2=1, 2-1=1. answer = 3. 
- The second line: 4-3=1, 3-3=0, 3-1=2. The answer is 3. 
- The third line: 4-3=1, 3-4=-1, Chef can't move to the first number here. Therefore, the answer is -1. 
- The fourth line: 5-5=0, 5-2=3, 2-1=1. The answer is 4.

```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 = 'inp = input().split()\n\nrow_n,col_m,operation_p = int(inp[0]),int(inp[1]),int(inp[2])\n\noperations = []\n\n#---------------Getting Operations-------------\nfor i in range(operation_p):\n inp = input().split()\n row,col = int(inp[0]),int(inp[1])\n operations.append([row,col])\noperations.sort()\n#------------------------------------------------\n\ni=0\nsol_index = 1\nwhile i<operation_p:\n if operations[i][0]==sol_index:\n  #Logic\n  sol = col_m-1\n  o = {}\n  while (i<operation_p and operations[i][0]==sol_index ):\n   e = operations[i][1]\n   if e in o:\n    o[e]+=1\n   else:\n    o[e]=1\n   i+=1\n  #print o\n  for e in o:\n   if o[e]>1:\n    # print "O[e] > 1"\n    if e+1<=col_m:\n     # print("e+1 <= col_m")\n     if e+1 in o:\n      #print "e+1 in o"\n      if o[e]-o[e+1]>1:\n       sol = -1\n       break\n     else:\n      sol = -1\n      break\n  if sol != -1:\n   if 1 in o:\n    sol-=o[1]\n   if col_m in o:\n    sol+=o[col_m]\n  print(sol)\n  sol_index+=1\n else:\n  print(col_m-1)\n  sol_index+=1\nwhile sol_index <= row_n:\n print(col_m-1)\n sol_index+=1'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13741