#include <iostream>
#include <vector>

int calc(std::vector<std::vector<int>> &obstacleGrid)
{

    int R = obstacleGrid.size();
    int C = obstacleGrid[0].size();

    // If the starting cell has an obstacle, then simply return as there would be
    // no paths to the destination.
    if (obstacleGrid[0][0] == 1)
    {
        return 0;
    }

    // Number of ways of reaching the starting cell = 1.
    obstacleGrid[0][0] = 1;

    // Filling the values for the first column
    for (int i = 1; i < R; i++)
    {
        obstacleGrid[i][0] = (obstacleGrid[i][0] == 0 && obstacleGrid[i - 1][0] == 1) ? 1 : 0;
    }

    // Filling the values for the first row
    for (int i = 1; i < C; i++)
    {
        obstacleGrid[0][i] = (obstacleGrid[0][i] == 0 && obstacleGrid[0][i - 1] == 1) ? 1 : 0;
    }

    // Starting from cell(1,1) fill up the values
    // No. of ways of reaching cell[i][j] = cell[i - 1][j] + cell[i][j - 1]
    // i.e. From above and left.
    for (int i = 1; i < R; i++)
    {
        for (int j = 1; j < C; j++)
        {
            if (obstacleGrid[i][j] == 0)
            {
                obstacleGrid[i][j] = obstacleGrid[i - 1][j] + obstacleGrid[i][j - 1];
            }
            else
            {
                obstacleGrid[i][j] = 0;
            }
        }
    }

    // Return value stored in rightmost bottommost cell. That is the destination.
    return obstacleGrid[R - 1][C - 1];
}

int main()
{
    int row, column, bs;
    std::cin >> row >> column >> bs;

    std::vector<std::vector<int>> grid(row, std::vector<int>(column, 0));

    for (int i = 0; i < bs; i++)
    {
        int r, c;
        std::cin >> r >> c;
        grid[r][c] = 1;
    }

    std::cout << calc(grid);

    return 0;
}