Let $N$ be a positive integer. We have a $(2N+1)\times (2N+1)$ grid where rows are numbered $0$ through $2N$ and columns are also numbered $0$ through $2N$. Let $(i,j)$ denote the square at Row $i$ and Column $j$.

We have one white pawn, which is initially at $(0, N)$. Also, we have $M$ black pawns, the $i$-th of which is at $(X_i, Y_i)$.

When the white pawn is at $(i, j)$, you can do one of the following operations to move it:

-   If $0\leq i\leq 2N-1$, $0 \leq j\leq 2N$ hold and $(i+1,j)$ does not contain a black pawn, move the white pawn to $(i+1, j)$.
-   If $0\leq i\leq 2N-1$, $0 \leq j\leq 2N-1$ hold and $(i+1,j+1)$ does contain a black pawn, move the white pawn to $(i+1,j+1)$.
-   If $0\leq i\leq 2N-1$, $1 \leq j\leq 2N$ hold and $(i+1,j-1)$ does contain a black pawn, move the white pawn to $(i+1,j-1)$.

You cannot move the black pawns.

Find the number of integers $Y$ such that it is possible to have the white pawn at $(2N, Y)$ by repeating these operations.

Input is given from Standard Input in the following format:

```
$N$ $M$
$X_1$ $Y_1$
$:$
$X_M$ $Y_M$
```

Print the answer.

### Constraints

-   $1 \leq N \leq 10^9$
-   $0 \leq M \leq 2\times 10^5$
-   $1 \leq X_i \leq 2N$
-   $0 \leq Y_i \leq 2N$
-   $(X_i, Y_i) \neq (X_j, Y_j)$ $(i \neq j)$
-   All values in input are integers.

3

We can move the white pawn to (4,0), (4,1), and (4,2), as follows:

• (0,2)→(1,1)→(2,1)→(3,1)→(4,2)
• (0,2)→(1,1)→(2,1)→(3,1)→(4,1)
• (0,2)→(1,1)→(2,0)→(3,0)→(4,0)

On the other hand, we cannot move it to (4,3) or (4,4). Thus, we should print 3.

0

We cannot move the white pawn from (0,1).