human
Can you solve the below in PYTHON3?
You are given a matrix, consisting of n rows and m columns. The rows are numbered top to bottom, the columns are numbered left to right.

Each cell of the matrix can be either free or locked.

Let's call a path in the matrix a staircase if it: 

  * starts and ends in the free cell.

  * visits only free cells.

  * has one of the two following structures: 
    1. the second cell is 1 to the right from the first one, the third cell is 1 to the bottom from the second one, the fourth cell is 1 to the right from the third one, and so on.

    2. the second cell is 1 to the bottom from the first one, the third cell is 1 to the right from the second one, the fourth cell is 1 to the bottom from the third one, and so on. 


In particular, a path, consisting of a single cell, is considered to be a staircase.

Here are some examples of staircases:


Initially all the cells of the matrix are free.

You have to process q queries, each of them flips the state of a single cell. So, if a cell is currently free, it makes it locked, and if a cell is currently locked, it makes it free.

Print the number of different staircases after each query. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.

Input

The first line contains three integers n, m and q (1 ≤ n, m ≤ 1000.
1 ≤ q ≤ 10^4) — the sizes of the matrix and the number of queries.

Each of the next q lines contains two integers x and y (1 ≤ x ≤ n.
1 ≤ y ≤ m) — the description of each query.

Output

Print q integers — the i-th value should be equal to the number of different staircases after i queries. Two staircases are considered different if there exists such a cell that appears in one path and doesn't appear in the other path.

Examples

Input

2 2 8
1 1
1 1
1 1
2 2
1 1
1 2
2 1
1 1

Output

5
10
5
2
5
3
1
0

Input

3 4 10
1 4
1 2
2 3
1 2
2 3
3 2
1 3
3 4
1 3
3 1

Output

49
35
24
29
49
39
31
23
29
27

Input

1000 1000 2
239 634
239 634

Output

1332632508
1333333000

=====
What is a Python code to print the number of different staircases after each query?
assistant
#Here is Python code to print the number of different staircases after each query.

import collections
import math
import itertools
import  sys
import time
import random

def fast_input():
    return sys.stdin.readline().strip()

input = fast_input

def ints_input():
    return list(int(x) for x in input().split())

def int_input():
    return int(input())


def solve():
    n, m, q = ints_input()
    free = [[1 for _ in range(m)] for _ in range(n)]
    staircases_on_diagonal = [0 for _ in range(-m, n)]
    total_non_singleton_staircases = 0
    total_singleton_staircases = n * m
    def calculate_staircases(x):
        if x <= 0:
            start_point = (0, -x)
        else:
            start_point = (x, 0)
        staircases = 0
        if start_point[0] > 0:
            running = free[start_point[0]-1][start_point[1]]
        else:
            running = 0
        while True:
            # print(x, start_point, running)
            # Update running
            if free[start_point[0]][start_point[1]]:
                running += 1
            else:
                running = 0
            if running > 1:
                staircases += running - 1
            # Move right
            start_point = (start_point[0], start_point[1] + 1)
            if start_point[1] >= m:
                break
            if free[start_point[0]][start_point[1]]:
                running += 1
            else:
                running = 0
            if running > 1:
                staircases += running - 1
            # Move down
            start_point = (start_point[0] + 1, start_point[1])
            if start_point[0] >= n:
                break
        return staircases
    for x in range(-(m-1), n):
        stairs = calculate_staircases(x)
        staircases_on_diagonal[x] = stairs
        total_non_singleton_staircases += stairs
    for _ in range(q):
        x, y = ints_input()
        x = x-1
        y = y-1
        free[x][y] = 1 - free[x][y]
        total_singleton_staircases += (1 if free[x][y] else -1)
        for diff in (x - y, x + 1 - y):
            if diff < -(m - 1) or diff >= n:
              ...