/*

     题意：有n个商店排成一条直线，有一些商店有先后顺序，问从0出发走到n+1最少的步数

     贪心：对于区间被覆盖的点只进行一次计算，还有那些要往回走的区间步数*2，再加上原来最少要走n+1步就是答案了

     详细解释：http://blog.csdn.net/u013625492/article/details/45640735

 */

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 using namespace std;

 const int MAXN = 1e3 + ;

 const int INF = 0x3f3f3f3f;

 struct A

 {

     int l, r;

 }a[MAXN];

 struct B

 {

     int l, r;

 }b[MAXN];

 bool cmp(A x, A y)    {return x.l < y.l;}

 int main(void)        //UVALive 6834 Shopping

 {

 //    freopen ("C.in", "r", stdin);

     int n, m;

     while (scanf ("%d%d", &n, &m) == )

     {

         int cnt = ;

         for (int i=; i<=m; ++i)

         {

             int u, v;    scanf ("%d%d", &u, &v);

             if (u > v)    continue;

             a[++cnt].l = u;    a[cnt].r = v;

         }

         sort (a+, a++cnt, cmp);

         int l = a[].l;    int r = a[].r;    int tot = ;

         for (int i=; i<=cnt; ++i)        //区间覆盖

         {

             if (r < a[i].l)

             {

                 b[++tot].l = l;    b[tot].r = r;

                 l = a[i].l;    r = a[i].r;

             }

             else    r = max (r, a[i].r);

             if (i == cnt)    {b[++tot].l = l;    b[tot].r = r;}

         }

         int ans = n + ;

         for (int i=; i<=tot; ++i)    ans +=  * (b[i].r - b[i].l);

         printf ("%d\n", ans);

     }

     return ;

 }