In particular $m\geq 3$, since $(0,m)\in A(m,n)$ or when $n<m$ the height of $\tilde{P}$ is at most $m-1$. Hence $$\tilde{P}=\cdots 1\underbrace{00\cdots 00}_{r}\underbrace{1010\cdots 1010}_{t}\underbrace{11\cdots 11}_{n},\ \ r\geq 2,\ \ t\in 2{\mathbb Z}_+.$$ If $t\neq 0$, we can set $$\tilde{P}=\cdots 1\underbrace{00\cdots 00}_{r-1}100\underbrace{10\cdots 1010}_{t-2}\underbrace{11\cdots 11}_{n},$$ which is admissible and has the desired property. If $t=0$, we set $$\bar{P}=\cdots 1\underbrace{11\cdots 11}_{N_{0,1}}\underbrace{00\cdots 00}_{r-1}\underbrace{11\cdots 11}_{n-2N_{0,1}}0\underbrace{11\cdots 11}_{N_{0,1}},$$ where $N_{0,1}=n_1+\delta_{n_0,m-1}$. We get $$\text{comaj}(\bar{P})=\text{comaj}(\tilde{P})-(n+r)+(n+r-N_{0,1})+(N_{0,1}+1)=\text{comaj}(\tilde{P})+1.$$

(0,0) grid (19,7); (0,0) – (3,3) – (6,0) – (9,3)–(12,0)–(19,7); (9,3) – (11,5) – (13,3)–(16,6)–(17,5); at (8,2) [$\ell$]{}; at (12,2) [$r$]{}; at (10,4) [$a$]{}; at (15,5) [$b$]{};

Again, we must prove that $\bar{P}$ is admissible. Recall that $r<m$, since $N=m-1$ or $(0,m)\in A(n,m)$. There are two new peaks $z_1$ and $z_2$ in the path $\bar{P}$, where $$z_1=(s-n-r+N_{0,1},r+N_{0,1}),\ \ z_2=(n-N_{0,1}-1,n-N_{0,1}+1).$$ We know that $$(n-N_{0,1}-1,n-N_{0,1}+1)\in A(n,m)$$ and hence the peak $z_2$ does not violate property . If $r+N_{0,1}<m$, there is nothing to check for $z_1$ and otherwise we have $(r+N_{0,1}+\delta_{n_0,m-1}-m,r+N_{0,1})\in A(m,n)$ and the claim follows with $$\text{ht}_{z'}(\bar{P})\geq N_{0,1}+1>r+N_{0,1}+\delta_{n_0,m-1}-m,\ \ \text{ $\forall z'$ with }\ell_{z'}(\bar{P})\geq s-n-r+N_{0,1}.$$ Hence in the case when $n\neq 0$ we have that $\mathcal{V}^{1\rightarrow m}_{s,n}(q)$ forms an interval of consecutive integers. If $n=0$ and $m=2$, the statement is clear since $\mathcal{D}_{2,0}(s)=\{P\}$. Similarly there is nothing 