def compute_product(i):
    """Compute product of floor(i^(1/k)) for all k >= 1"""
    if i == 1:
        return 1
    
    product = 1
    k = 1
    
    # For i <= 10^18, we need at most ~60 iterations
    while k <= 63:
        # Binary search for floor(i^(1/k))
        low, high = 1, min(i, 10**9)
        
        while low < high:
            mid = (low + high + 1) // 2
            if mid ** k <= i:
                low = mid
            else:
                high = mid - 1
        
        kth_root = low
        
        if kth_root == 1:
            # All further roots are 1
            break
        
        product *= kth_root
        k += 1
    
    return product


def solve(N):
    MOD = 998244353
    result = 0
    
    for i in range(1, N + 1):
        result = (result + compute_product(i)) % MOD
    
    return result


N = int(input())
print(solve(N))