#include <bits/stdc++.h>
#define ll              long long
#define pii             pair<int,int>
#define pll             pair<ll,ll>
#define rep(i,m,n)      for(int i=(m); i<=(n); i++)
#define reb(i,m,n)      for(int i=(m); i>=(n); i--)
#define iter(id, v)     for(auto id : v)
#define fs              first
#define se              second
#define MP              make_pair
#define pb              push_back
#define bit(msk, i)     ((msk >> i) & 1)
#define SZ(v)           (ll)v.size()
#define ALL(v)          v.begin(),v.end()

using namespace std;

mt19937_64 rd(chrono :: steady_clock :: now ().time_since_epoch().count());
ll Rand (ll l, ll r) { return uniform_int_distribution<ll> (l, r) (rd); }

const int N = 5e3 + 7;
const int Mod = 1e9 + 7; ///loonf mod sai
const ll INF = 1e18;
const ll BASE = 137;
const int szBL = 320;

int n, m;
vector<int> ke[N], ke2[N];
pii edges[N];

namespace sub3 {
    int vis[N], vis2[N];

    void dfs (int u) {
        vis[u] = 1;
        iter (&v, ke[u]) if (!vis[v]) dfs(v);
    }

    void dfs2 (int u) {
        vis2[u] = 1;
        iter (&v, ke2[u]) if (!vis2[v]) dfs2(v);
    }

    bool check (int K, int pos) {
        rep (i, 1, n) ke[i].clear(), ke2[i].clear();
        rep (i, 1, K) {
            ke[edges[i].fs].pb(edges[i].se), ke2[edges[i].se].pb(edges[i].fs);
        }
        rep (i, 1, n) vis[i] = vis2[i] = 0;
        dfs(pos);
        dfs2(pos);
        int res = 0, res2 = 0;
        rep (i, 1, n) {
            res += vis[i], res2 += vis2[i];
        }
        return res + res2 - 1 == n;
    }

    void solution () {
        rep (i, 1, n) {
            int lf = 1, rt = m;
            while (lf < rt) {
                int mid = lf + rt >> 1;
                if (check(mid, i)) rt = mid;
                else lf = mid + 1;
            }
            if (check(lf, i)) cout << lf <<" ";
            else cout << -1 <<" ";
        }
    }
}

namespace sub4 {
    void solution() {
            
    }
}

void solution() {
    cin >> n >> m;
    rep (i, 1, m) {
        int u, v;
        cin >> u >> v;
        edges[i] = {u, v};
    }
    if (n <= 1000 && m <= 4000) sub3 :: solution();
//    else sub4 :: solution();

}

#define file(name) freopen(name".inp","r",stdin); \
freopen(name".out","w",stdout);
int main () {
//    file("c");
    ios_base :: sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int num_Test = 1;
//    cin >> num_Test;
    while (num_Test--)
        solution();
}
/*
no bug +8
chu y break hay return se lam hong logic
xet transition cua i va i + 1
construct ket qua
chu y truong hop : KHONG CO GI

ko làm được

hướng 1: đổi hướng làm
hướng 2: đưa ra nhận xét

tim mo hinh bai toan sau khi doc de

trung ten bien trong ham gan nhat co the dan den sai

5 6
3 2 3
4 2 1
5 3 9
1 3 5
1 4 2
2 3 1


*/