Problem

A graph is given in which N vertices, each numbered from 1 to N, are connected by N-1 undirected edges. For each vertex, output the shortest number of steps to start from that vertex and visit all vertices.

However, one step is to follow one side from one vertex and move to another vertex.

Constraints

* 2 ≤ N ≤ 105
* 1 ≤ ui, vi ≤ N (1 ≤ i ≤ N-1)
* ui ≠ vi
* The graph given is concatenated

Input

The input is given in the following format.


N
u1 v1
..
..
..
uN−1 vN−1


The first line is given one integer N.
In the following N-1 line, the integers ui and vi representing the vertex numbers at both ends of the i-th side are given on the i-th line, separated by blanks.

Output

Output the shortest number of steps to visit all vertices starting from vertex i on the i-th line from vertex 1 to vertex N.

Examples

Input

2
1 2


Output

1
1


Input

6
1 2
1 3
3 4
3 5
5 6


Output

7
6
8
7
7
6
#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <queue>
#include <algorithm>
#include <utility>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define INF_LL (ll)1e18
#define INF (int)1e9
#define REP(i, n) for(int i = 0;i < (n);i++)
#define FOR(i, a, b) for(int i = (a);i < (b);i++)
#define all(x) x.begin(),x.end()
using ll = long long;
using PII = pair<int, int>;

template<typename A, typename B>inline void chmin(A &a, B b){if(a > b) a = b;}
template<typename A, typename B>inline void chmax(A &a, B b){if(a < b) a = b;}

class Union_find{
private:
	vector<int> par;
	vector<int> rank;
	int n;

public:
	Union_find(int a){
		n = a;
		for(int i = 0;i < n;i++){
			par.push_back(i);
			rank.push_back(0);
		}
	}

	int find(int x){
		if(par[x] == x){
			return x;
		}else{
			return par[x] = find(par[x]);
		}
	}

	void unite(int x, int y){
		x = find(x);
		y = find(y);
		if(x == y) return;

		if(rank[x] < rank[y]){
			par[x] = y;
		}else{
			par[y] = x;
			if(rank[x] == rank[y]) rank[x]++;
		}
	}

	bool same(int x, int y){
		return find(x) == find(y);
	}
};

class LazySegTree{
private:
	int n;
	vector<ll> node, lazy;
	vector<bool> lazyFlag;
public:
	LazySegTree(vector<ll> v){
		int sz = v.size();
		n = 1; while(n < sz) n *= 2;
		node.resize(2*n-1, 0);
		lazy.resize(2*n-1, 0);
		lazyFlag.resize(2*n-1, false);
		REP(i, sz) node[i+n-1] = v[i];
		for(int i = n-2;i >= 0;i--) node[i] = node[2*i+1]+node[2*i+2];
	}

	void eval(int k, int l, int r){
		if(lazyFlag[k]){
			node[k] += lazy[k] * (r-l);
			if(r-l > 1){
				lazy[2*k+1] += lazy[k];
				lazy[2*k+2] += lazy[k];
				lazyFlag[2*k+1] = lazyFlag[2*k+2] = true;
			}
			lazy[k] = 0;
			lazyFlag[k] = false;
		}
	}

	void add(int a, int b, ll x, int k=0, int l=0, int r=-1){
		if(r < 0) r = n;
		eval(k, l, r);
		if(b <= l || r <= a) return;

		if(a <= l && r <= b){
			lazy[k] += x;
			lazyFlag[k] = true;
			eval(k, l, r);
		}else{
			add(a, b, x, k*2+1, l, (l+r)/2);
			add(a, b, x, k*2+2, (l+r)/2, r);
			node[k] = node[k*2+1] + node[k*2+2];
		}
	}

	ll query(int a, int b, int k=0, int l=0, int r=-1){
		if(r < 0) r = n;
		eval(k, l, r);

		if(b <= l || r <= a) return 0;
		if(a <= l && r <= b) return node[k];

		return query(a, b, k*2+1, l, (l+r)/2) + query(a, b, k*2+2, (l+r)/2, r);
	}
};

class Bucket{
private:
	int n, bs, bn;
	vector<ll> data;
	vector<ll> bucket, lazy;
	vector<bool> lazyFlag;
public:
	Bucket(vector<ll> v, int sz){
		data = v;
		n = v.size();
		bs = sz;
		bn = (n + bs - 1) / bs;
		data.resize(bn*bs, INF_LL);
		bucket.assign(bn, INF_LL);
		lazy.assign(bn, 0);
		lazyFlag.assign(bn, 0);
		REP(i, bn){
			ll minx = INF_LL;
			REP(j, bs){
				chmin(minx, data[i*bs+j]);
			}
			bucket[i] = minx;
		}
	}

	void eval(int k){
		if(lazyFlag[k]){
			lazyFlag[k] = false;
			FOR(i, bs*k, bs*(k+1)){
				data[i] = lazy[k];
			}
			lazy[k] = 0;
		}
	}

	void update(int s, int t, int x){
		REP(k, bn){
			int l = k*bs, r = (k+1)*bs;
			if(r <= s || t <= l) continue;
			if(s <= l && r <= t){
				lazyFlag[k] = true;
				bucket[k] = x;
				lazy[k] = x;
			}else{
				eval(k);
				FOR(i, max(s, l), min(t, r)){
					data[i] = x;
				}
				bucket[k] = INF_LL;
				FOR(i, l, r){
					chmin(bucket[k], data[i]);
				}
			}
		}
	}

	ll query(int s, int t){
		ll res = INF_LL;
		REP(k, bn){
			int l = k*bs, r = (k+1)*bs;
			if(r <= s || t <= l) continue;
			if(s <= l && r <= t){
				chmin(res, bucket[k]);
			}else{
				eval(k);
				FOR(i, max(s, l), min(t, r)){
					chmin(res, data[i]);
				}
			}
		}
		return res;
	}

};

int N;
vector<int> G[114514];
int d[114514] = {};
int ans[114514] = {};

void dfs1(int v, int p){
	REP(i, G[v].size()){
		if(G[v][i] == p) continue;
		dfs1(G[v][i], v);
		d[v] = max(d[v], d[G[v][i]]+1);
	}
}

void dfs2(int v, int p, int dp){
	vector<PII> dd;
	dd.emplace_back(0, -1);

	REP(i, G[v].size()){
		if(G[v][i] == p) dd.emplace_back(dp + 1, p);
		else dd.emplace_back(d[G[v][i]]+1, G[v][i]);
	}
	sort(dd.rbegin(), dd.rend());
	ans[v] = dd[0].first;
	REP(i, G[v].size()){
		if(G[v][i] == p) continue;
		dfs2(G[v][i], v, dd[G[v][i] == dd[0].second].first);
	}

}

int main(void){
	cin.tie(0);
	ios::sync_with_stdio(false);
	
	cin >> N;
	REP(i, N-1){
		int u, v; cin >> u >> v; u--; v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs1(0, -1);
	dfs2(0, -1, 0);
	REP(i, N){
		cout << (N-1)*2-ans[i] << endl;
	}
}