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 comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N = 500005, M = N * 4;
struct Gragh {
  int cnt, y[M], nxt[M], fst[N];
  void clear() {
    cnt = 1;
    memset(fst, 0, sizeof fst);
  }
  void add(int a, int b) { y[++cnt] = b, nxt[cnt] = fst[a], fst[a] = cnt; }
} g, g2;
int n, m;
int dfn[N], low[N], inst[N], st[N], vis[N], id[N], Time, top, tot;
int used[M];
int Fa[N];
vector<int> cir[N], son[N];
void Tarjan_Prepare() {
  Time = top = tot = 0;
  memset(id, 0, sizeof id);
  memset(st, 0, sizeof st);
  memset(dfn, 0, sizeof dfn);
  memset(low, 0, sizeof low);
  memset(vis, 0, sizeof vis);
  memset(inst, 0, sizeof inst);
  memset(used, 0, sizeof used);
}
void Tarjan(int x) {
  dfn[x] = low[x] = ++Time;
  vis[x] = inst[x] = 1;
  st[++top] = x;
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    if (used[i / 2]) continue;
    used[i / 2] = 1;
    if (!vis[g.y[i]]) {
      Tarjan(g.y[i]);
      low[x] = min(low[x], low[g.y[i]]);
    } else if (inst[g.y[i]])
      low[x] = min(low[x], low[g.y[i]]);
  }
  if (dfn[x] == low[x]) {
    tot++;
    id[st[top]] = tot;
    inst[st[top]] = 0;
    while (st[top--] != x) {
      id[st[top]] = tot;
      inst[st[top]] = 0;
    }
  }
}
void Get_cir(int x) {
  if (vis[x]) return;
  vis[x] = 1;
  cir[id[x]].push_back(x);
  for (int i = g.fst[x]; i; i = g.nxt[i]) {
    int y = g.y[i];
    if (id[y] == id[x]) Get_cir(y);
  }
}
void build(int x, int pre) {
  int ID = id[x];
  Fa[ID] = pre;
  cir[ID].clear();
  Get_cir(x);
  for (int i = 0; i < cir[ID].size(); i++) {
    int u = cir[ID][i];
    son[u].clear();
    for (int j = g.fst[u]; j; j = g.nxt[j]) {
      int v = g.y[j];
      if (id[v] == ID || id[v] == pre) continue;
      son[u].push_back(v);
      build(v, ID);
    }
  }
}
int Deep[N];
int spDeep[N], usDeep[N];
int Lmax[N], Rmax[N];
void Get_Deep(int x) {
  int ID = id[x];
  for (int i = 0; i < cir[ID].size(); i++) {
    int u = cir[ID][i], v;
    Deep[u] = 0;
    for (int j = 0; j < son[u].size(); j++) {
      v = son[u][j];
      Get_Deep(v);
      Deep[u] = max(Deep[u], Deep[v] + 1);
    }
  }
  int n = cir[ID].size() - 1;
  Lmax[0] = Rmax[n + 1] = 0;
  for (int i = 1; i <= n; i++) Lmax[i] = max(Lmax[i - 1], Deep[cir[ID][i]] + i);
  for (int i = n; i >= 1; i--)
    Rmax[i] = max(Rmax[i + 1], Deep[cir[ID][i]] + (n - i + 1));
  int v = 1e9;
  for (int i = 1; i <= n + 1; i++) v = min(v, max(Lmax[i - 1], Rmax[i]));
  Deep[x] = max(usDeep[x] = Deep[x], spDeep[x] = v);
}
int Far[N];
int val[N * 3];
int Lv[N * 3], Rv[N * 3];
struct Monotone_Queue {
  int head, tail, q[N * 3], d[N * 3];
  void clear() { head = 1, tail = 0; }
  int front() { return head <= tail ? q[head] : -1e9; }
  void push(int v, int _d) {
    while (head <= tail && q[tail] <= v) tail--;
    q[++tail] = v, d[tail] = _d;
  }
  void pop(int _d) {
    while (head <= tail && d[head] < _d) head++;
  }
  int Next_front(int _d) {
    if (head > tail) return -1e9;
    if (d[head] != _d) return q[head];
    return head == tail ? -1e9 : q[head + 1];
  }
  void print() {
    printf("size=%d, sit=:", tail - head + 1);
    for (int i = head; i <= tail; i++) printf("(%d,%d)  ", q[i], d[i]);
    puts("");
  }
} QL, QR;
vector<int> LRmax[N];
void Solve(int x, int far) {
  int ID = id[x];
  Far[x] = far;
  int n = cir[ID].size();
  for (int i = 1; i < n; i++) val[i] = Deep[cir[ID][i]];
  val[0] = max(usDeep[x], Far[x]);
  for (int i = n; i < n * 3; i++) val[i] = val[i % n];
  for (int i = 0; i < n * 3; i++) {
    Lv[i] = val[i] + n * 3 - i;
    Rv[i] = val[i] + i + 1;
  }
  QL.clear(), QR.clear();
  int Lp = 2, Rp = n + 1;
  for (int i = Lp; i <= n; i++) QL.push(Lv[i], i);
  for (int i = n + 1; i < n * 2; i++) {
    int result = 1e9;
    int v1 = n * 3 - i, v2 = i + 1;
    while (Rp < i) {
      Lp++, QL.pop(Lp);
      Rp++;
      if (Rp != i) QR.push(Rv[Rp], Rp);
    }
    while (QL.Next_front(Lp) - v1 > max(QR.front(), Rv[Rp + 1]) - v2) {
      result = min(result, QL.front() - v1);
      Lp++, QL.pop(Lp);
      Rp++, QR.push(Rv[Rp], Rp);
    }
    result = min(result, QL.front() - v1);
    result = min(result, max(QR.front(), Rv[Rp + 1]) - v2);
    QL.push(Lv[i], i);
    QR.pop(i + 2);
    Far[cir[ID][i % n]] = result;
  }
  for (int i = 0; i < n; i++) {
    int u = cir[ID][i];
    int m = son[u].size();
    Lmax[0] = Rmax[m + 1] = 0;
    for (int j = 1; j <= m; j++)
      Lmax[j] = max(Lmax[j - 1], Deep[son[u][j - 1]] + 1);
    for (int j = m; j >= 1; j--)
      Rmax[j] = max(Rmax[j + 1], Deep[son[u][j - 1]] + 1);
    LRmax[u].clear();
    for (int j = 0; j < m; j++) LRmax[u].push_back(max(Lmax[j], Rmax[j + 2]));
    for (int j = 0; j < m; j++)
      Solve(son[u][j], max(spDeep[u], max(Far[u], LRmax[u][j])) + 1);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  g.clear();
  for (int i = 1, a, b; i <= m; i++) {
    scanf("%d%d", &a, &b);
    g.add(a, b);
    g.add(b, a);
  }
  Tarjan_Prepare();
  for (int i = 1; i <= n; i++)
    if (!vis[i]) Tarjan(i);
  memset(vis, 0, sizeof vis);
  build(1, 0);
  Get_Deep(1);
  Solve(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", max(Far[i], Deep[i]));
  return 0;
}