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#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;
}
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