You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multiple edges between any pair of vertices. Graph can be disconnected.

You should assign labels to all vertices in such a way that:

  * Labels form a valid permutation of length n — an integer sequence such that each integer from 1 to n appears exactly once in it. 
  * If there exists an edge from vertex v to vertex u then labelv should be smaller than labelu. 
  * Permutation should be lexicographically smallest among all suitable. 



Find such sequence of labels to satisfy all the conditions.

Input

The first line contains two integer numbers n, m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105).

Next m lines contain two integer numbers v and u (1 ≤ v, u ≤ n, v ≠ u) — edges of the graph. Edges are directed, graph doesn't contain loops or multiple edges.

Output

Print n numbers — lexicographically smallest correct permutation of labels of vertices.

Examples

Input

3 3
1 2
1 3
3 2


Output

1 3 2 


Input

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


Output

4 1 2 3 


Input

5 4
3 1
2 1
2 3
4 5


Output

3 1 2 4 5 
#include <bits/stdc++.h>
using namespace std;
struct Matrix {
  vector<vector<int>> data;
  int r, c;
  Matrix(int row, int col, bool identity = false) : r(row), c(col) {
    data.assign(row, vector<int>(col, 0));
    if (identity) {
      for (int i = 0; i < r; ++i) {
        data[i][i] = 1;
      }
    }
  }
  Matrix operator*(Matrix& other) {
    int m = r, n = c, p = other.c;
    Matrix res(m, p);
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < p; ++j) {
        for (int k = 0; k < n; ++k) {
          res.data[i][j] += data[i][k] * other.data[k][j];
        }
      }
    }
    return res;
  }
};
vector<int> extendGcd(int a, int b) {
  if (b == 0) {
    return {a, 1, 0};
  } else {
    vector<int> tmp = extendGcd(b, a % b);
    return {tmp[0], tmp[2], tmp[1] - (a / b) * tmp[2]};
  }
}
Matrix matrix_power(Matrix base, long long exp) {
  int n = base.r;
  Matrix res(n, n, true);
  while (exp) {
    if (exp & 1) {
      res = res * base;
    }
    base = base * base;
    exp >>= 1;
  }
  return res;
}
long long power_mod(long long base, int exp) {
  long long res = 1;
  while (exp) {
    if (exp & 1) res = res * base % 1000000007;
    base = base * base % 1000000007;
    exp >>= 1;
  }
  return res;
}
long long power(long long base, int exp) {
  long long res = 1;
  while (exp) {
    if (exp & 1) res *= base;
    base *= base;
    exp >>= 1;
  }
  return res;
}
long long inv(long long a) { return power_mod(a, 1000000007 - 2); }
long long fac[100010];
void initFac() {
  fac[0] = 1;
  for (int i = 1; i < 100010; ++i) {
    fac[i] = i * fac[i - 1] % 1000000007;
  }
}
long long Combine(long long a, long long b) {
  return (fac[a] * inv(fac[b]) % 1000000007) * inv(fac[a - b]) % 1000000007;
}
int n, m;
int label[100010];
unordered_map<int, unordered_set<int>> pre;
unordered_map<int, int> out;
priority_queue<int> pq;
int main() {
  cin >> n >> m;
  memset(label, -1, sizeof(label));
  for (int i = 0; i < m; ++i) {
    int v, u;
    cin >> v >> u;
    pre[u].insert(v);
    ++out[v];
    out[u];
  }
  for (int i = 1; i <= n; ++i) {
    if (out[i] == 0) {
      pq.push(i);
      out.erase(i);
    }
  }
  for (int i = n; i >= 1; --i) {
    int ind = pq.top();
    label[ind] = i;
    pq.pop();
    for (int j : pre[ind]) {
      if (--out[j] == 0) {
        pq.push(j);
        out.erase(j);
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (i > 1) cout << " ";
    cout << label[i];
  }
  return 0;
}
