Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.

However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.

In how many ways he can do it? Two ways are called different if the remaining common part is different.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.

Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).

Output

Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.

Examples

Input

3 2
1 2 3
2 3 1


Output

4


Input

5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5


Output

5


Input

2 2
1 2
2 1


Output

2

Note

In the first example, all possible common parts are [1], [2], [3] and [2, 3].

In the second and third examples, you can only leave common parts of length 1.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.*;
import java.io.*;
public class codeforces 
{ 
   static class Student{
       int x,y;
        Student(int x,int y){
            this.x=x;
            this.y=y;
        }
    }
    static int prime[];
    static void sieveOfEratosthenes(int n) 
    { 
        // Create a boolean array "prime[0..n]" and initialize 
        // all entries it as true. A value in prime[i] will 
        // finally be false if i is Not a prime, else true. 
        int pos=0;
        prime= new int[n+1]; 
        for(int p = 2; p*p <=n; p++) 
        { 
            // If prime[p] is not changed, then it is a prime 
            if(prime[p] == 0)
            { 
                // Update all multiples of p
                prime[p]=p;
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = p; 
            } 
        } 
    }
static    class Sortbyroll implements Comparator<Student> 
{ 
    // Used for sorting in ascending order of 
    // roll number 
    public int compare(Student c, Student b) 
    { 
        return c.y-b.y;
    } 
} 
    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
    static class Edge{
        int a,b;
        Edge(int a,int b){
            this.a=a;
            this.b=b;
        }
    }
    static int parent[];
    static int rank[];
    static int seg[];
    public static void main(String[] args){
        //long sum=0;
        FastReader sc=new FastReader();
        int n,m,i,j,k,pos;
        long ans=0,c;
        n=sc.nextInt();
        m=sc.nextInt();
        int a[][]=new int[m][n];
        for(i=0;i<m;i++){
            for(j=0;j<n;j++){
                a[i][j]=sc.nextInt();
            }
        }
        ArrayList<HashMap<Integer,Integer>>ar=new ArrayList<>();
        for(i=0;i<m;i++){
            ar.add(new HashMap<>());
        }
        for(i=0;i<m;i++){
            ar.get(i).put(0,0);
            for(j=0;j<n;j++){
                ar.get(i).put(a[i][j],j+1);
            }
        }
        c=0;
        i=0;
        while(i<n){
            pos=0;
            c=1;
            j=i+1;
            while(j<n&&pos==0){
                for(k=0;k<m;k++){
                    if(ar.get(k).get(a[0][j])!=1+ar.get(k).get(a[0][j-1])){
                        pos=1;
                        break;
                    }
                }
                if(pos==1)
                    break;
                ++c;
                ++j;
            }
            ans=ans+c*(c+1)/2;
            i=j;
        }
        System.out.println(ans);
    }
    static int query(int s,int e,int l,int r,int in){
        if(s>e||s>r|e<l)
            return 0;
        if(s>=l&&e<=r)
            return seg[in];
        return query(s,(s+e)/2,l,r,2*in+1)+query((s+e)/2+1,e,l,r,2*in+2);
    }
    static void update(int s,int e,int l,int r,int in,int val){
        if(s>e||s>r|e<l)
            return ;
        if(s>=l&&e<=r)
        {seg[in]+=val;
         return;
        }
        update(s,(s+e)/2,l,r,2*in+1,val);
        update((s+e)/2+1,e,l,r,2*in+2,val);
        seg[in]=seg[2*in+1]+seg[2*in+2];
    }
 /*static long power(long x, long y, long p) 
{ 
    long res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if ((y & (long)1)%2!=0) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;   
    } 
    return res%p; 
} */
   static int find(int x) 
    { 
        // Finds the representative of the set 
        // that x is an element of 
        while(parent[x]!=x) 
        { 
            // if x is not the parent of itself 
            // Then x is not the representative of 
            // his set,
            x=parent[x];
  
            // so we recursively call Find on its parent 
            // and move i's node directly under the 
            // representative of this set 
        } 
  
        return x; 
    }
    static void union(int x, int y) 
    { 
        // Find representatives of two sets 
        int xRoot = find(x), yRoot = find(y); 
  
        // Elements are in the same set, no need 
        // to unite anything. 
        if (xRoot == yRoot) 
            return; 
  
         // If x's rank is less than y's rank 
        if (rank[xRoot] < rank[yRoot]) 
  
            // Then move x under y  so that depth 
            // of tree remains less 
            parent[xRoot] = yRoot; 
  
        // Else if y's rank is less than x's rank 
        else if (rank[yRoot] < rank[xRoot]) 
  
            // Then move y under x so that depth of 
            // tree remains less 
            parent[yRoot] = xRoot; 
  
        else // if ranks are the same 
        { 
            // Then move y under x (doesn't matter 
            // which one goes where) 
            parent[yRoot] = xRoot; 
  
            // And increment the the result tree's 
            // rank by 1 
            rank[xRoot] = rank[xRoot] + 1; 
        } 
    } 
    static long gcd(long a, long b) 
    { 
        if (a == 0) 
            return b; 
        return gcd(b % a, a); 
    }
}