Problem:
Score : 500 points

Problem Statement

There is a rooted tree (see Notes) with N vertices numbered 1 to N.
Each of the vertices, except the root, has a directed edge coming from its parent.
Note that the root may not be Vertex 1.

Takahashi has added M new directed edges to this graph.
Each of these M edges, u \rightarrow v, extends from some vertex u to its descendant v.

You are given the directed graph with N vertices and N-1+M edges after Takahashi added edges.
More specifically, you are given N-1+M pairs of integers, (A_1, B_1), ..., (A_{N-1+M}, B_{N-1+M}), which represent that the i-th edge extends from Vertex A_i to Vertex B_i.

Restore the original rooted tree.

Notes

For "tree" and other related terms in graph theory, see the article in Wikipedia, for example.

Constraints

3 \leq N

1 \leq M

N + M \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

If i \neq j, (A_i, B_i) \neq (A_j, B_j).

The graph in input can be obtained by adding M edges satisfying the condition in the problem statement to a rooted tree with N vertices.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_{N-1+M} B_{N-1+M}

Output

Print N lines.
In the i-th line, print 0 if Vertex i is the root of the original tree, and otherwise print the integer representing the parent of Vertex i in the original tree.

Note that it can be shown that the original tree is uniquely determined.

Sample Input 1

3 1
1 2
1 3
2 3

Sample Output 1

0
1
2

The graph in this input is shown below:

It can be seen that this graph is obtained by adding the edge 1 \rightarrow 3 to the rooted tree 1 \rightarrow 2 \rightarrow 3.

Sample Input 2

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

Sample Output 2

6
4
2
0
6
2

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
 * じょえチャンネル
 * 高評価・チャンネル登録よろしくおねがいします！
 * https://www.youtube.com/channel/UCRXsI3FL_kvaVL9zoolBfbQ
 */

#include <bits/stdc++.h>

//#pragma GCC target("avx2")
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")

#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)

//here!!!
// define int long long !!!!!

#define int long long

// define int long long !!!!!


//#define mod (int)((1e9)+7)
constexpr int mod = 998244353ll;

#define intt long long
#define itn long long
#define P pair<int,int>

#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)

#define ALL(v) v.begin(),v.end()

using namespace std;

//Library
//モッドパウ
inline int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y & 1) {
            res *= x;
            res %= m;
        }
        x = x * x % m;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}


//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}


//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}


template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}


#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",i)
#define fendl printf("\n")

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    //    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}
//Library-End


#define vecin(v) for(int i=0;i<v.size();i++)scanf("%lld",&v[i]);
#define vecout(v) {for(int i=0;i<(int)v.size();i++)printf("%lld ",v[i]);printf("\n");}

//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理
    
    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }
    
public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
            function<T(T, T)> _update)
    : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }
    
    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }
    
    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }
    
    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
    vector<int> ookisa;
    
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        ookisa=vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
            ookisa[i]=1;
        }
    }
    
    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }
    
    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
            ookisa[y]+=ookisa[x];
            ookisa[x]=0;
        }
        else {
            par[y] = x;
            ookisa[x]+=ookisa[y];
            ookisa[y]=0;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }
    int size(int i){
        i=find(i);
        return ookisa[i];
    }
    bool same(int x, int y){
        return find(x) == find(y);
    }
    
};

//Union-Find-End

int perm[1000005];
void init_perm() {
    perm[0] = 1;
    REP(i, 1000000) {
        perm[i] = perm[i - 1] * i % mod;
    }
}

int nCk(int x, int y) {
    return perm[x] * modpow(perm[x - y], mod - 2) % mod * modpow(perm[y], mod - 2) % mod;
}

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}

inline string stringmax(string& x,string& y){
    if (x.size()>y.size()) {
        return x;
    }
    if (x.size()<y.size()) {
        return y;
    }
    rep(i,x.size()){
        if (x[i]>y[i]) {
            return x;
        }
        if (x[i]<y[i]) {
            return y;
        }
    }
    return x;
}

#define endl "\n" //interactive の時に注意！！！



int n,m,a[100004],b[100004],p,ans[100004];
vector<itn> road[100004];
vector<int> topo;
bool went[100004],oya[100004],did[100004];

void dfs(int x){
    went[x]=1;
    rep(i,road[x].size()){
        if (went[road[x][i]]) {
            continue;
        }
        dfs(road[x][i]);
    }
    topo.emplace_back(x);
}
signed main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    cin>>n>>m;
    rep(i,n+m-1){
        cin>>a[i]>>b[i];
        road[a[i]].emplace_back(b[i]);
        oya[b[i]]=1;
    }
    REP(i,n){
        if (!oya[i]) {
            p=i;
            break;
        }
    }
    dfs(p);
    reverse(ALL(topo));
    ans[topo[0]]=0;
    for(auto& i:topo){
        for(auto& j:road[i]){
            if (did[j]) {
                continue;
            }
            ans[j]=i;
        }
    }
    REP(i,n)cout<<ans[i]<<endl;
}