------
## 最终结论
```python
def minCost(n, cuts):
    # Add the boundaries of the stick to the cuts list
    cuts = sorted(cuts + [0, n])
    
    # Initialize a 2D DP array
    dp = [[0] * (len(cuts)) for _ in range(len(cuts))]
    
    # Fill the DP table
    for length in range(2, len(cuts)):  # start from length 2 because we need at least two points to make a cut
        for i in range(len(cuts) - length):
            j = i + length
            dp[i][j] = float('inf')
            for k in range(i + 1, j):
                cost = cuts[j] - cuts[i] + dp[i][k] + dp[k][j]
                dp[i][j] = min(dp[i][j], cost)
    
    return dp[0][-1]

# Example usage:
n1, cuts1 = 7, [1, 3, 4, 5]
print(minCost(n1, cuts1))  # Output: 16

n2, cuts2 = 9, [5, 6, 1, 4, 2]
print(minCost(n2, cuts2))  # Output: 22
```