Actually visible points

An $N$ dimensional coordinate system holds several points. They are described by $M$ natural numbers $X_1, X_2, \dots, X_M$, and the set $S_i$ described by $X_i$ is

$S_i = \{(x_1, x_2, \dots, x_N) \mid 1 \le x_1 \le x_2 \le \dots \le x_N = X_i\}$

where $x_1, x_2, \dots, x_N$ are all natural numbers.

Let $S = S_1 \cup S_2 \cup \dots \cup S_M$. The points that lie in the coordinate system are exactly the points of $S$.

Count the points visible from the origin $(0, 0, \dots, 0)$. A point $p$ is visible when the segment joining the origin and $p$ contains no point of $S$ other than $p$ itself. A lattice coordinate on that segment that does not belong to $S$ blocks nothing.

The first line contains two natural numbers $N$ and $M$ separated by a space. ($2 \le N \le 100000$, $1 \le M \le 100000$)

Each of the next $M$ lines contains one natural number $X_i$. ($1 \le X_i \le 100000$) The $X_i$ are pairwise distinct.

Print the number of points visible from the origin, modulo $1000000007$.

For $N = 2$ and $X$ equal to 1, 2, 3, 4, the points of $S$ are $(1,1)$, $(1,2)$, $(2,2)$, $(1,3)$, $(2,3)$, $(3,3)$, $(1,4)$, $(2,4)$, $(3,4)$, $(4,4)$. Six of them are visible: $(1,1)$, $(1,2)$, $(1,3)$, $(2,3)$, $(1,4)$, $(3,4)$. The point $(4,4)$ is hidden by $(3,3)$, which lies on the segment and belongs to $S$, and $(2,4)$ is hidden by $(1,2)$.

With $X$ equal to 4 alone the answer changes. Then $S$ is only $(1,4)$, $(2,4)$, $(3,4)$, $(4,4)$, and $(1,2)$ is missing from $S$, so all four points are visible.