The first two subtasks can be solved using Dynamic Programming. Every point ~(x,y)~ can only be reached from points ~(x-1,y)~ or ~(x,y-1)~. Therefore, the amount of paths you can take to ~(x,y)~ is equal to the paths to ~(x-1,y)~ and ~(x,y-1)~ summed together. Store the paths to ~(x,y)~ in a ~2\text{D}~ array, and recursively sum to find the total amount of paths to ~(m,n)~

The last subtask requires knowledge of properties of Pascal's triangle. Notice that by the above sum property, the grid follows a pattern similar to Pascal's Triangle. Another property is that the ~k~th number in the ~n~th row (the first row is row ~0~) is equal to ~\binom{n}{k} = \frac{n!} {(k! \ (n-k)!)}~. A proof of this property can be found in the footnote. By this property, the amount of paths can be found by computing ~\binom{m+n}{m} = \frac{(m+n)!}{m! \ \ n!}~

Thus we are presented with a new problem, how do we calculate the value effectively? Taking the factorials and then dividing would cause an overflow as ~n,m \leq 10^7~. The solution requires taking mod after every factorial. We can perform division with mod by using the property of modular inverse. Given ~a \ (\text{mod } N)~, if ~\text{GCD} \ (a,N) = 1~ (the numbers are coprime), then there exists an integer ~x < N~ such that ~ax \equiv 1 \ \ (\text{mod } N)~. ~x~ is the modular inverse. Dividing by ~a~ is equivalent to multiplying by ~x~, the modular inverse. Because ~10^9+7~ is a prime number, ~(a,10^9+7)~ will always be coprime and there always exists a modular inverse.

One way to find the modular inverse is by looping through all values ~\text{mod } N~. However, we have to loop through at most ~N~ operations to find the inverse. We can do one better by using Fermat's little theorem, which states that for a prime ~p~, ~a^p \equiv a \ (\text{mod }p)~, and ~a^{p-1} \equiv 1 \ (\text{mod }p)~ if ~(a,p)~ coprime. By Fermat's little theorem, we get that ~a \cdot a^{10^9+5} \equiv 1 \ (\text{mod }p)~, which means ~a^{10^9+5}~ is the inverse of ~a \ (\text{mod }N)~.

But this is still not fast enough!!!!! We will have to do at most ~10^9+5~ operations if we do the exponential normally! Next, we use binary exponentiation, which is done by computing the mods of ~a~ to the exponent of a power of ~2~. For example if we want to compute ~a^{14}~, we just multiply together ~a^8, a^4~ and ~a^2~. Here is the implementation of binary exponentiation in pseudocode:

exp = 1
x = 1000000005
while x > 0{
  if x & 1 == 1{
    exp = exp * b % 1000000007;
  }
  a = a*a % 1000000007;
  x = x >> 1;
}

return exp

Footnote To prove the Pascal triangle property, we can use Pascal's formula, which states that ~\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}~. Try proving that for yourself! It can also be solved by considering a set of ~m~ up operations and ~n~ right operations. The amount of paths is equal to the amount of distinct permutations of the set, which also gives ~\binom{m+n}{m}~