We have a grid with N rows and N columns. The cell at the i-th row and j-th column is denoted (i, j).

Initially, M of the cells are painted black, and all other cells are white. Specifically, the cells (a_1, b_1), (a_2, b_2), ..., (a_M, b_M) are painted black.

Snuke will try to paint as many white cells black as possible, according to the following rule:

* If two cells (x, y) and (y, z) are both black and a cell (z, x) is white for integers 1≤x,y,z≤N, paint the cell (z, x) black.



Find the number of black cells when no more white cells can be painted black.

Constraints

* 1≤N≤10^5
* 1≤M≤10^5
* 1≤a_i,b_i≤N
* All pairs (a_i, b_i) are distinct.

Input

The input is given from Standard Input in the following format:


N M
a_1 b_1
a_2 b_2
:
a_M b_M


Output

Print the number of black cells when no more white cells can be painted black.

Examples

Input

3 2
1 2
2 3


Output

3


Input

2 2
1 1
1 2


Output

4


Input

4 3
1 2
1 3
4 4


Output

3
#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;

typedef long long ll;
const int MAXN=1e5+5;

int n,m,flag;
int cnt[3],ncnt,ecnt,c[MAXN];
bool vis[MAXN],evis[MAXN];
struct node{
    int v,w,id;
    node(int _v,int _w,int _id):v(_v),w(_w),id(_id){}
};
vector<node> E[MAXN];

void dfs(int u){
    vis[u]=1;
    ncnt++;
    cnt[c[u]]++;
    for(int i=0;i<(int)E[u].size();i++){
        int v=E[u][i].v,w=E[u][i].w,id=E[u][i].id;
        if(evis[id]) continue; evis[id]=1; ecnt++;
        if(!vis[v]){
            c[v]=(c[u]+w+3)%3;
            dfs(v);
        }
        else if(c[v]!=(c[u]+w+3)%3) flag=1;
    }
}

int main(){

	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
	    int u,v;
        scanf("%d%d",&u,&v);
        E[u].push_back(node(v,1,i));
        E[v].push_back(node(u,-1,i));
	}
	ll ans=0;
    for(int i=1;i<=n;i++){
        if(vis[i]) continue;
        flag=ncnt=ecnt=cnt[0]=cnt[1]=cnt[2]=0;
        dfs(i);
        if(flag) ans+=1ll*ncnt*ncnt;
        else if(!cnt[0]||!cnt[1]||!cnt[2]) ans+=ecnt;
        else ans+=1ll*cnt[0]*cnt[1]+1ll*cnt[1]*cnt[2]+1ll*cnt[2]*cnt[0];
    }
    printf("%lld",ans);

	fclose(stdin);
	fclose(stdout);
}
