if(g[i][k] and g[k][j]) g[i][j]=1;

/*#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<bitset>

#include<vector>

using namespace std;

inline int read()

{

    int x=0,w=0;char c=getchar();

    while(!isdigit(c))w|=c=='-',c=getchar();

    while(isdigit(c))x=(x<<3)+(x<<1)+(c^48),c=getchar();

    return w?-x:x;

}

const int maxn=1e3+10;

int n,m;

bitset<maxn>g[maxn];

int main()

{

    n=read(),m=read();

    for(int i=1;i<=m;i++)

        g[read()][read()]=1;

    for(int k=1;k<=n;k++)

        for(int i=1;i<=n;i++)

            for(int j=1;j<=n;j++)

            if(g[i][k] & g[k][j])g[i][j]=1;

    for(int i=1;i<=n;i++) if(g[i][i]){

        printf("-1\n");return 0;

    }

    int ans=0;

    for(int i=1;i<=n;i++)

        for(int j=1;j<=n;j++)

            if(!g[i][j] and !g[j][i])ans++;

    printf("%d",(ans-n)/2);

    return 0;

}*/

//上面是邻接矩阵

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

#define R register

using namespace std;

inline int read()

{

    int x=0,w=0;char c=getchar();

    while(!isdigit(c))w|=c=='-',c=getchar();

    while(isdigit(c))x=(x<<3)+(x<<1)+(c^48),c=getchar();

    return w?-x:x;

}

const int maxn=1100;

int head[2][maxn],to[2][maxn],nxt[2][maxn],ecnt,n,m;

inline void addedge(int from,int too)

{

    to[0][++ecnt]=too,nxt[0][ecnt]=head[0][from],head[0][from]=ecnt;

    to[1][ecnt]=from,nxt[1][ecnt]=head[1][too],head[1][too]=ecnt;

}

bool vis[maxn][maxn];

int main()

{

    n=read(),m=read();

    int ans=0;

    for(R int x,y,i=1;i<=m;i++)

    {

        x=read(),y=read();

        if(!vis[x][y])

            addedge(x,y),ans++,vis[x][y]=1;

    }

    for(R int u,v,k=1;k<=n;k++)

        for(R int i=head[1][k];i;i=nxt[1][i])

        {

            u=to[1][i];

            for(R int j=head[0][k];j;j=nxt[0][j])

            {

                v=to[0][j];

                if(vis[u][v])continue;

                vis[u][v]=1;

                ans++;

                addedge(u,v);

            }

        }

    printf("%d",n*(n-1)/2-ans);

    return 0;

}