To solve this problem, we need to determine the number of valid paths from the top-left corner to the bottom-right corner of an n x n grid, where each path can only move right or down, and at any point, the number of right moves must be greater than or equal to the number of down moves.

Approach

The problem can be efficiently solved using Catalan numbers. Here's the key insight:

• A valid path from (0,0) to (n-1, n-1) with the given constraint is equivalent to a Dyck path of length 2*(n-1). The number of such paths is given by the (n-1)-th Catalan number.
• The formula for the k-th Catalan number is: ( Ck = \frac{1}{k+1} \binom{2k}{k} ), where ( k = n-1 ). Thus, the number of valid paths for an n x n grid is ( C{n-1} = \frac{1}{n} \binom{2(n-1)}{n-1} ).

To compute this efficiently modulo ( 10^9+7 ):

1. Precompute factorials up to ( 2*(1000-1) = 1998 ) (since n can be up to 1000).
2. Precompute inverse factorials using Fermat's Little Theorem (since ( 10^9+7 ) is a prime number).
3. Precompute inverses of numbers up to 1000.

Solution Code

MOD = 10**9 + 7
max_n = 1000
max_fact = 2 * (max_n - 1)  # 1998

# Precompute factorials modulo MOD
fact = [1] * (max_fact + 1)
for i in range(1, max_fact + 1):
    fact[i] = fact[i-1] * i % MOD

# Precompute inverse factorials modulo MOD using Fermat's Little Theorem
inv_fact = [1] * (max_fact + 1)
inv_fact[max_fact] = pow(fact[max_fact], MOD-2, MOD)
for i in range(max_fact - 1, -1, -1):
    inv_fact[i] = inv_fact[i+1] * (i+1) % MOD

# Precompute inverses of numbers up to max_n modulo MOD
inv = [1] * (max_n + 1)
for i in range(2, max_n + 1):
    inv[i] = pow(i, MOD-2, MOD)

# Read input and compute result
n = int(input())
if n == 1:
    print(1)
else:
    k = n - 1
    # Calculate combination C(2k, k)
    comb = fact[2*k] * inv_fact[k] % MOD
    comb = comb * inv_fact[k] % MOD
    # Calculate Catalan number C_k = comb * inv[k+1]
    catalan = comb * inv[k+1] % MOD
    print(catalan)

Explanation

1. Factorial Precomputation: We compute factorials up to 1998 to quickly calculate combinations.
2. Inverse Factorial Precomputation: Using Fermat's Little Theorem, we compute the inverse of each factorial, which helps in calculating combinations efficiently.
3. Inverse of Numbers: We compute the inverse of each number up to 1000 to apply the Catalan number formula directly.
4. Result Calculation: For a given n, we compute the (n-1)-th Catalan number using the precomputed values, which gives the number of valid paths.

This approach ensures that we handle large values efficiently and correctly using modular arithmetic, making it suitable for the given constraints. The time complexity for precomputation is O(2000), which is negligible, and each query is answered in O(1) time.

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