You Are Given a Tree

D. You Are Given a Tree

time limit per test

7 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

A tree is an undirected graph with exactly one simple path between each pair of vertices. We call a set of simple paths kk-valid if each vertex of the tree belongs to no more than one of these paths (including endpoints) and each path consists of exactly kk vertices.

You are given a tree with nn vertices. For each kk from 11 to nn inclusive find what is the maximum possible size of a kk-valid set of simple paths.

Input

The first line of the input contains a single integer nn (2≤n≤1000002≤n≤100000) — the number of vertices in the tree.

Then following n−1n−1 lines describe the tree, each of them contains two integers vv, uu (1≤v,u≤n1≤v,u≤n) — endpoints of the corresponding edge.

It is guaranteed, that the given graph is a tree.

Output

Output nn numbers, the ii-th of which is the maximum possible number of paths in an ii-valid set of paths.

Examples

input

Copy

7
1 2
2 3
3 4
4 5
5 6
6 7

output

Copy

7
3
2
1
1
1
1

input

Copy

6
1 2
2 3
2 4
1 5
5 6

output

Copy

6
2
2
1
1
0

Note

One way to achieve the optimal number of paths for the second sample is illustrated in the following picture:

#include <bits/stdc++.h>

#define lowbit(x) x&(-x)
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 108;
const ll mod = 1e9 + 7;
int ooo = 800;

struct node{
    int to,nx;
}o[maxn];

int n,tot,head[maxn],cal[maxn],res[maxn];
int path;

inline void add_edge(int u,int v){
    o[++tot].to=v;
    o[tot].nx=head[u];
    head[u]=tot;
}

inline void dfs(int cur,int fa,int chose){
    cal[cur]=0;
    bool flag=false;
    for(register int i=head[cur];i;i=o[i].nx){
        int to=o[i].to;
        if(to==fa)continue;
        dfs(to,cur,chose);
        if(cal[cur]+cal[to]>=chose-1)flag=true;
        cal[cur]=max(cal[cur],cal[to]);
    }
    if(flag||chose==1){
        ++path;
        cal[cur]=0;
    }
    else{
        ++cal[cur];
    }
}

inline void solve(int l,int r,int res_l,int res_r){
    if(l>r)return;
    if(res_l==res_r){
        for(register int i=l;i<=r;++i){
            res[i]=res_l;
        }
        return;
    }
    int mid=l+r>>1;
    path=0;
    dfs(1,0,mid);
    res[mid]=path;
    solve(l,mid-1,res[mid],res_r);
    solve(mid+1,r,res_l,res[mid]);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    scanf("%d",&n);
    for(register int i=1,u,v;i<n;++i){
        scanf("%d%d",&u,&v);
        add_edge(u,v);
        add_edge(v,u);
    }
    solve(1,n,0,n);
    for(register int i=1;i<=n;++i){
        printf("%d\n",res[i]);
    }
    return 0;
}