Is it possible to determine the given matrix is positive semidefinite under these conditions?

Suppose I have a $2^n$ by $2^n$ symmetric matrix M. I know the following facts are true about $M$.

1. The diagonal of M is $n+1$, which is strictly larger than any other non-diagonal entry.

2. The sum of each row of the matrix M is exactly $2^n$

3. The value of non-diagonal entry cannot be smaller than $-n+1$

4. Each row contains the same elements (but the order is different so that $M$ is symmetric)

I really hope to conclude $M$ is positive semidefinite, but I have to admit this may not be true. I know the fact that if each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row, $M$ would be positive definite. However, we cannot use here because of fact 2. On the other hand, fact 2 implies $M$ has eigenvector 1 with eigenvalue $2^n$ and there cannot be too many negative entries in $M$. I wonder if these conditions can be sufficient for me to draw such a conclusion.

Courant fischer theorem seems helpful here since we can express the smallest eigenvalue as $\min_{v \perp 1} \frac{v^{T}Mv}{v^{T}v}$

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