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 mm neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 11 to nn. 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 nn and mm (1≤n≤1000001≤n≤100000, 1≤m≤101≤m≤10) — the number of suspects and the number of asked neighbors.

Each of the next mm lines contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n). It is guaranteed that these integers form a correct permutation (that is, each number from 11 to nn 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][1], [2][2], [3][3] and [2,3][2,3].

In the second and third examples, you can only leave common parts of length 11.

 

题意：

给你k个互不相同的1~n的全排列，

求这k个排列有多少个公共子序列。

思路：

这题关键的一点是全排列的性质，每一个数都仅且出现1次。

利用这个性质，我们可以建立一个pre数组，a[i] [x ] 表示的是在第i个全排列中 x这个数前面的数。

那么我们只需要从一个全排列下手，来求他的子序列是否也是其他全部排列的子序列，

利用一个cnt变量来维护当前已经满足条件的子序列长度。

如果当前的全排列x前面的数y，其他的全排列中x前面的数也是y，那么cnt++，否则把cnt赋值为1，（一个数也是满足条件的子序列）

（上面用到的是组合数学的性质，即3个长度的序列有 3+2+1个子序列 那么维护的时候加起来也是1+2+3 ）

 

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int pre[maxn];
int a[12][100005];
int n,k;
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>n>>k;
    repd(i,1,k)
    {
        repd(j,1,n)
        {
            cin>>pre[j];// pre数组中最终存的是最后一行的值
            a[i][pre[j]]=pre[j-1]; //  a[i][j]  表示第i个全排列中，j之前的数
        }
    }
    ll cnt=0;
    ll ans=0;
    repd(j,1,n)
    {
        int isok=1;
        repd(i,1,k-1)
        {
            if(a[i][pre[j]]!=pre[j-1])// 判断第i行中是否存在最后一行的第j 位和第j-1 位
            {
                isok=0;
                break;
            }
        }
        if(isok)
        {
            cnt++;
        }else
        {
            cnt=1;
        }
        ans+=cnt;
    }
    cout<<ans<<endl;


    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}