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 JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

import static java.util.Collections.reverse;

/**
 * @author Don Li
 */
public class CactusToTree {
    
    int N = (int) 5e5 + 10;
    
    int n, m;
    int[][][] G;
    
    int timer = 0;
    int[] tin = new int[N], low = new int[N];
    boolean[] bridge = new boolean[N * 2];
    
    int nc = 0;
    boolean[] vis = new boolean[N];
    int[] vertex_to_cycle = new int[N];
    List<Integer>[] cycle = new List[N];
    
    int[] par = new int[N]; 
    int[] len = new int[N]; 
    int[] L = new int[N];   
    int[] Z = new int[N];  
    
    void solve() {
        n = in.nextInt(); m = in.nextInt();
        int[] fr = new int[m], to = new int[m], wt = new int[m];
        for (int i = 0; i < m; i++) {
            fr[i] = in.nextInt() - 1;
            to[i] = in.nextInt() - 1;
            wt[i] = i;
        }
        G = build_graph(n, fr, to, wt);
        
        mark_bridges(0, -1);
        
        for (int i = 0; i < n; i++) {
            if (!vis[i]) {
                cycle[nc] = new ArrayList<>();
                dfs(i, nc++);
            }
        }
        
        create_spanning_tree_and_init_l(0);
        
        solve_cycle(0, -1, 0);
        
        for (int i = 0; i < n; i++) {
            if (i > 0) out.print(' ');
            out.print(Math.max(L[i], Z[i]));
        }
    }
    
    void solve_cycle(int u, int p, int carry) {
        L[u] = Math.max(L[u], carry);
        
        int id = vertex_to_cycle[u]; int k = cycle[id].size();
        for (int it = 0; it < 2; it++) {
            PriorityQueue<int[]> qu = new PriorityQueue<>((a, b) -> {
                if (a[0] != b[0]) return -Integer.compare(a[0], b[0]);
                return -Integer.compare(a[1], b[1]);
            });
            for (int i = 0; i < 2 * k; i++) {
                int d = 2 * k - i;
                while (!qu.isEmpty() && qu.peek()[1] > d + k / 2) qu.poll();
                if (!qu.isEmpty()) Z[cycle[id].get(i % k)] = Math.max(Z[cycle[id].get(i % k)], qu.peek()[0] - d);
                qu.offer(new int[]{L[cycle[id].get(i % k)] + d, d});
            }
            reverse(cycle[id]);
        }
        
        for (int v : cycle[id]) {
            int fi = 0, se = 0;
            for (int[] t : G[v]) {
                int w = t[0], e = t[1];
                if (w != p && bridge[e]) {
                    if (fi < 1 + len[w]) {
                        se = fi;
                        fi = 1 + len[w];
                    } else if (se < 1 + len[w]) {
                        se = 1 + len[w];
                    }
                }
            }
            for (int[] t : G[v]) {
                int w = t[0], e = t[1];
                if (w != p && bridge[e]) {
                    int passValue = Math.max(carry, Z[v]);
                    if (1 + len[w] == fi) passValue = Math.max(passValue, se);
                    else passValue = Math.max(passValue, fi);
                    solve_cycle(w, v, passValue + 1);
                }
            }
        }
    }
    
    void create_spanning_tree_and_init_l(int s) {
        int[] qu = new int[n]; boolean[] inq = new boolean[n];
        int qh = 0, qt = 0;
        qu[qt++] = s;
        inq[s] = true;
        par[s] = -1;
        while (qh < qt) {
            int u = qu[qh++];
            for (int[] t : G[u]) {
                int v = t[0];
                if (!inq[v]) {
                    qu[qt++] = v;
                    inq[v] = true;
                    par[v] = u;
                }
            }
        }
        
        for (int i = n - 1; i >= 0; i--) {
            int u = qu[i];
            for (int[] t : G[u]) {
                int v = t[0], e = t[1];
                if (par[v] == u) {
                    len[u] = Math.max(len[u], len[v] + 1);
                    if (bridge[e]) L[u] = Math.max(L[u], len[v] + 1);
                }
            }
        }
    }
    
    void dfs(int u, int id) {
        vis[u] = true;
        vertex_to_cycle[u] = id;
        cycle[id].add(u);
        for (int[] t : G[u]) {
            int v = t[0], e = t[1];
            if (!bridge[e] && !vis[v]) {
                dfs(v, id);
            }
        }
    }
    
    void mark_bridges(int u, int p) {
        low[u] = tin[u] = ++timer;
        for (int[] t : G[u]) {
            int v = t[0], e = t[1];
            if (tin[v] == 0) {
                mark_bridges(v, u);
                low[u] = Math.min(low[u], low[v]);
                if (low[v] > tin[u]) bridge[e] = true;
            } else if (v != p) {
                low[u] = Math.min(low[u], tin[v]);
            }
        }
    }
    
    int[][][] build_graph(int n, int[] from, int[] to, int[] weight) {
        int[][][] G = new int[n][][];
        int[] cnt = new int[n];
        for (int i = 0; i < from.length; i++) {
            cnt[from[i]]++;
            cnt[to[i]]++;
        }
        for (int i = 0; i < n; i++) G[i] = new int[cnt[i]][];
        for (int i = 0; i < from.length; i++) {
            G[from[i]][--cnt[from[i]]] = new int[]{to[i], weight[i]};
            G[to[i]][--cnt[to[i]]] = new int[]{from[i], weight[i]};
        }
        return G;
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new CactusToTree().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}