Problem: The employees of the R1 company often spend time together: they watch football, they go camping, they solve contests. So, it's no big deal that sometimes someone pays for someone else.

Today is the day of giving out money rewards. The R1 company CEO will invite employees into his office one by one, rewarding each one for the hard work this month. The CEO knows who owes money to whom. And he also understands that if he invites person x to his office for a reward, and then immediately invite person y, who has lent some money to person x, then they can meet. Of course, in such a situation, the joy of person x from his brand new money reward will be much less. Therefore, the R1 CEO decided to invite the staff in such an order that the described situation will not happen for any pair of employees invited one after another.

However, there are a lot of employees in the company, and the CEO doesn't have a lot of time. Therefore, the task has been assigned to you. Given the debt relationships between all the employees, determine in which order they should be invited to the office of the R1 company CEO, or determine that the described order does not exist.

Input

The first line contains space-separated integers n and m <image> — the number of employees in R1 and the number of debt relations. Each of the following m lines contains two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi), these integers indicate that the person number ai owes money to a person a number bi. Assume that all the employees are numbered from 1 to n.

It is guaranteed that each pair of people p, q is mentioned in the input data at most once. In particular, the input data will not contain pairs p, q and q, p simultaneously.

Output

Print -1 if the described order does not exist. Otherwise, print the permutation of n distinct integers. The first number should denote the number of the person who goes to the CEO office first, the second number denote the person who goes second and so on.

If there are multiple correct orders, you are allowed to print any of them.

Examples

Input

2 1
1 2


Output

2 1 


Input

3 3
1 2
2 3
3 1


Output

2 1 3 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cout << to_string(H) << " | ";
  dbgm(T...);
}
const long long MOD = (int)1e9 + 7;
string ascii_lowercase = "abcdefghijklmnopqrstuvwxyz";
string ascii_uppercase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
struct hashLL {
  size_t operator()(long long x) const {
    x = (x ^ (x >> 30)) * UINT64_C(0xbf58476d1ce4e5b9);
    x = (x ^ (x >> 27)) * UINT64_C(0x94d049bb133111eb);
    x = x ^ (x >> 31);
    return x;
  }
};
long long inv(long long a, long long b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
const long long INF = 1e6;
unordered_map<int, int, custom_hash> mp;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long distsq(long long x1, long long y1, long long x2, long long y2) {
  return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
}
long long n, m;
unordered_map<long long, set<long long>> g;
const int N = 30004;
bool vis[N];
vector<int> v;
void dfs(int x) {
  if (vis[x]) return;
  vis[x] = true;
  for (auto nei : g[x]) {
    if (!vis[nei]) dfs(nei);
  }
  v.push_back(x);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  memset(vis, 0, sizeof(vis));
  long long a, b;
  for (int i = 0; i < (m); i++) {
    cin >> a >> b;
    g[b].emplace(a);
  }
  for (int i = 1; i <= (n); i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    cout << v[i] << " ";
  }
  return 0;
}