Problem: You are given a special connected undirected graph where each vertex belongs to at most one simple cycle.

Your task is to remove as many edges as needed to convert this graph into a tree (connected graph with no cycles). 

For each node, independently, output the maximum distance between it and a leaf in the resulting tree, assuming you were to remove the edges in a way that minimizes this distance.

Input

The first line of input contains two integers n and m (1 ≤ n ≤ 5⋅ 10^5), the number of nodes and the number of edges, respectively.

Each of the following m lines contains two integers u and v (1 ≤ u,v ≤ n, u ≠ v), and represents an edge connecting the two nodes u and v. Each pair of nodes is connected by at most one edge.

It is guaranteed that the given graph is connected and each vertex belongs to at most one simple cycle.

Output

Print n space-separated integers, the i-th integer represents the maximum distance between node i and a leaf if the removed edges were chosen in a way that minimizes this distance.

Examples

Input

9 10
7 2
9 2
1 6
3 1
4 3
4 7
7 6
9 8
5 8
5 9


Output

5 3 5 4 5 4 3 5 4


Input

4 4
1 2
2 3
3 4
4 1


Output

2 2 2 2

Note

In the first sample, a possible way to minimize the maximum distance from vertex 1 is by removing the marked edges in the following image:

<image>

Note that to minimize the answer for different nodes, you can remove different edges.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize(3)
using namespace std;
inline char Sc() {
  static const int LEN = 100000;
  static char Buf[LEN], *OP = Buf, *ED = Buf;
  if (OP == ED) {
    ED = (OP = Buf) + fread(Buf, 1, LEN, stdin);
    if (OP == ED) return -1;
  }
  return *OP++;
}
template <class T>
inline void Rd(T &x) {
  static char c;
  for (x = 0, c = Sc(); c < 48; c = Sc())
    ;
  for (; c > 47; c = Sc()) x = (x << 1) + (x << 3) + (c ^ 48);
}
int n, m;
struct Chart {
  int head[500005];
  Chart() { memset(head, -1, sizeof(head)); }
  struct edge {
    int to, nxt;
    edge(int _to = 0, int _nxt = 0) : to(_to), nxt(_nxt) {}
  } G[1000005 << 1];
  int tol;
  void addedge(int a, int b) {
    G[tol] = edge(b, head[a]);
    head[a] = tol++;
  }
  edge operator[](const int x) { return G[x]; }
} E, G;
int Dfn[500005], Low[500005], T;
int Col[500005], Cres;
int Stk[500005], top;
bool Vis[500005];
vector<int> P[500005];
void Tj(int x, int f) {
  Dfn[x] = Low[x] = ++T;
  Stk[++top] = x;
  Vis[x] = 1;
  for (int i = E.head[x]; ~i; i = E[i].nxt) {
    int y = E[i].to;
    if (y == f) continue;
    if (!~Dfn[y]) {
      Tj(y, x);
      Low[x] = min(Low[x], Low[y]);
    } else if (Vis[y])
      Low[x] = min(Low[x], Dfn[y]);
  }
  if (Low[x] == Dfn[x]) {
    Cres++;
    int y;
    do {
      y = Stk[top--];
      Col[y] = Cres;
      Vis[y] = 0;
      P[Cres].push_back(y);
    } while (x != y);
  }
}
int Q[500005];
bool Visroot[500005];
int Root[500005];
void BFS() {
  int L = 1, R = 1;
  Q[1] = 1;
  Vis[1] = 1;
  while (L <= R) {
    int x = Q[L++];
    if (!Visroot[Col[x]]) Visroot[Col[x]] = 1, Root[Col[x]] = x;
    for (int i = E.head[x]; ~i; i = E[i].nxt) {
      int y = E[i].to;
      if (Vis[y]) continue;
      G.addedge(x, y);
      Vis[y] = 1;
      Q[++R] = y;
    }
  }
}
int SLen[500005][2];
int ULen[500005];
int DLen[500005];
void dfs1(int x) {
  for (int i = G.head[x]; ~i; i = G[i].nxt) {
    int y = G[i].to;
    dfs1(y);
    int mx = SLen[y][0];
    if (mx + 1 > SLen[x][0]) {
      SLen[x][1] = SLen[x][0];
      SLen[x][0] = mx + 1;
    } else if (mx + 1 > SLen[x][1]) {
      SLen[x][1] = mx + 1;
    }
    if (Col[x] != Col[y]) {
      DLen[x] = max(DLen[x], mx + 1);
    }
  }
}
int dp[500005];
int Cir[500005];
int Ans[500005];
void Dp(int x) {
  int root = Root[x];
  int cnt = 0;
  int sz = (int)P[x].size();
  int pos;
  for (int i = (0), i_END_ = (sz - 1); i <= i_END_; ++i)
    if (P[x][i] == root) {
      pos = i;
      break;
    }
  for (int i = (pos), i_END_ = (sz - 1); i <= i_END_; ++i) Cir[++cnt] = P[x][i];
  for (int i = (0), i_END_ = (pos - 1); i <= i_END_; ++i) Cir[++cnt] = P[x][i];
  int L = 1, R = 0;
  int mid = sz / 2;
  Q[++R] = 1;
  dp[1] = max(ULen[root], DLen[root]);
  Ans[root] = ULen[root];
  for (int i = (2), i_END_ = (sz); i <= i_END_; ++i) {
    int id = Cir[i];
    while (L <= R && i - Q[L] > mid) L++;
    dp[i] = DLen[id];
    Ans[id] = dp[Q[L]] + i - Q[L];
    while (L <= R && dp[i] - i >= dp[Q[R]] - Q[R]) R--;
    Q[++R] = i;
  }
  for (int i = (2), i_END_ = (mid); i <= i_END_; ++i) {
    int id = Cir[i];
    while (L <= R && i + sz - Q[L] > mid) L++;
    if (L <= R) Ans[id] = max(Ans[id], dp[Q[L]] + i + sz - Q[L]);
  }
  for (int i = (2), i_END_ = (sz); i <= i_END_; ++i)
    if (sz - i + 2 > i) swap(Cir[i], Cir[sz - i + 2]);
  L = 1;
  R = 0;
  Q[++R] = 1;
  for (int i = (2), i_END_ = (sz); i <= i_END_; ++i) {
    int id = Cir[i];
    while (L <= R && i - Q[L] > mid) L++;
    dp[i] = DLen[id];
    Ans[id] = max(Ans[id], dp[Q[L]] + i - Q[L]);
    while (L <= R && dp[i] - i >= dp[Q[R]] - Q[R]) R--;
    Q[++R] = i;
  }
  for (int i = (2), i_END_ = (mid); i <= i_END_; ++i) {
    int id = Cir[i];
    while (L <= R && i + sz - Q[L] > mid) L++;
    if (L <= R) Ans[id] = max(Ans[id], dp[Q[L]] + i + sz - Q[L]);
  }
}
void dfs2(int x) {
  if (Root[Col[x]] == x) Dp(Col[x]);
  for (int i = G.head[x]; ~i; i = G[i].nxt) {
    int y = G[i].to;
    if (Col[y] != Col[x]) {
      if (SLen[x][0] == SLen[y][0] + 1) {
        ULen[y] = max(Ans[x], SLen[x][1]) + 1;
      } else {
        ULen[y] = max(Ans[x], SLen[x][0]) + 1;
      }
    }
    dfs2(y);
  }
}
void Init() {
  for (int i = (1), i_END_ = (n); i <= i_END_; ++i) Dfn[i] = -1;
  Tj(1, 0);
  BFS();
}
void Solve() {
  dfs1(1);
  dfs2(1);
  for (int i = (1), i_END_ = (n); i <= i_END_; ++i)
    printf("%d\n", max(Ans[i], SLen[i][0]));
}
int main() {
  Rd(n);
  Rd(m);
  int a, b;
  for (int i = (1), i_END_ = (m); i <= i_END_; ++i) {
    Rd(a);
    Rd(b);
    E.addedge(a, b);
    E.addedge(b, a);
  }
  Init();
  Solve();
  return 0;
}