import sys

MAX_R = 2000001 MOD = 10**9 + 7 FACT = [1] * MAX_R

def power(a, b): res = 1 a %= MOD while b > 0: if b % 2 == 1: res = (res * a) % MOD a = (a * a) % MOD b //= 2 return res

def modInverse(n): return power(n, MOD - 2)

def precompute_factorials(): for i in range(2, MAX_R): FACT[i] = (FACT[i-1] * i) % MOD

def combinations(n, k): if k < 0 or k > n: return 0 if k == 0 or k == n: return 1 if k > n // 2: k = n - k

numerator = FACT[n]
denominator = (FACT[k] * FACT[n - k]) % MOD

return (numerator * modInverse(denominator)) % MOD

def solve_matrix_tracing(M, N): R = M + N - 2 K = M - 1

if R < 0:
    return 0

return combinations(R, K)

if name == 'main': precompute_factorials()

try:
    t_line = sys.stdin.readline()
    if not t_line:
        sys.exit(0)

    t = int(t_line.strip())

    for _ in range(t):
        try:
            line = sys.stdin.readline().rstrip().split()
            if not line:
                continue

            M = int(line[0])
            N = int(line[1])

            result = solve_matrix_tracing(M, N)

            print(result)

        except EOFError:
            break
        except Exception as e:
            continue

except Exception as e:
    pass

As others have been saying, the answer numerically should be (m+n-2) choose (m-1) (mod 10^9 + 7). Of course, the problem is the time limit, which you will see if you try this method. Here are the two insights that I needed to solve this:

1. since the answer is mod p, make your own factorial function that takes mod p for every successive multiplication
2. Use FLT and fast exponentiation mod p to divide by (m-1)! and (n-1)! instead of calculating them and then dividing

Dive into the neon grid, dodging digital projectiles like a Moto X3M rider evading obstacles. Matrix Tracing: a mind-bending game of reflexes and precision. Each pulse of light tests your skill, demanding split-second decisions.

For those interested, the technique needed to solve this exercises is close to the implementation of math.comb in the Python standard library (albeit using modulo 2⁶⁴ which comes for free if using unsigned integers). You can watch Raymond Hettinger's keynote "Numerical Marvels Inside Python" for more explanations.

TLE'd on this but i was thinking somewhere along the lines of the followng: nc = int(math.factorial(m)//(math.factorial(r)*math.factorial(m-r))) where r in range(n)