#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

typedef long long ll;

int const maxn=1e5+,maxm=5e5+;

int n,m,hd[maxn],ct,siz[maxn],dfn[maxn],low[maxn],tim;

ll ans[maxn];

struct N{

    int to,nxt;

    N(int t=,int n=):to(t),nxt(n) {}

}ed[maxm<<],edge[maxm<<];

void add(int x,int y){ed[++ct]=N(y,hd[x]); hd[x]=ct;}

void tarjan(int x,int f)

{

    dfn[x]=low[x]=++tim;

    ll t=; siz[x]=;//注意t是ll，否则下面算给ans时爆int

    for(int i=hd[x],u;i;i=ed[i].nxt)

    {

        if((u=ed[i].to)==f)continue;

        if(!dfn[u])

        {

            tarjan(u,x); siz[x]+=siz[u];

            low[x]=min(low[x],low[u]);

            if(low[u]>=dfn[x])ans[x]+=siz[u]*t,t+=siz[u];//t是割点的子树大小

        }

        else low[x]=min(low[x],dfn[u]);

    }

    ans[x]+=t*(n-t-);

}

int main()

{

    scanf("%d%d",&n,&m);

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

    {

        scanf("%d%d",&x,&y);

        add(x,y); add(y,x);

    }

    tarjan(,);

    for(int i=;i<=n;i++)printf("%lld\n",(ans[i]+n-)*);

    return ;

}