/*

51 nod 1405 树的距离之和

problem:

给定一棵无根树，假设它有n个节点，节点编号从1到n, 对于每个i求所有点到i的和。

solve:

假设已经知道了所有点到u的和, 对于它右儿子v的和,可以发现增加了Size[左子树]条uv边.减少了Size[v]条uv边

所以先求出所有点到1的和,然后可以推出其它所有点

hhh-2016/09/13-20:15:59

*/

#pragma comment(linker,"/STACK:124000000,124000000")

#include <algorithm>

#include <iostream>

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <vector>

#include <math.h>

#include <queue>

#include <set>

#include <map>

#define lson  i<<1

#define rson  i<<1|1

#define ll long long

#define clr(a,b) memset(a,b,sizeof(a))

#define scanfi(a) scanf("%d",&a)

#define scanfs(a) scanf("%s",a)

#define scanfl(a) scanf("%I64d",&a)

#define scanfd(a) scanf("%lf",&a)

#define key_val ch[ch[root][1]][0]

#define eps 1e-7

#define inf 0x3f3f3f3f3f3f3f3f

using namespace std;

const int maxn = 100110;

template<class T> void read(T&num)

{

    char CH;

    bool F=false;

    for(CH=getchar(); CH<'0'||CH>'9'; F= CH=='-',CH=getchar());

    for(num=0; CH>='0'&&CH<='9'; num=num*10+CH-'0',CH=getchar());

    F && (num=-num);

}

int stk[70], tp;

template<class T> inline void print(T p)

{

    if(!p)

    {

        puts("0");

        return;

    }

    while(p) stk[++ tp] = p%10, p/=10;

    while(tp) putchar(stk[tp--] + '0');

    putchar('\n');

}

ll ans[maxn];

struct node

{

    int to,next;

}edge[maxn <<2];

int tot,n;

int head[maxn];

int Size[maxn];

ll f[maxn];

void add_edge(int u,int v)

{

    edge[tot].to = v,edge[tot].next = head[u],head[u] = tot ++;

}

void dfs(int u,ll len,int pre)

{

    f[u] = len;

    Size[u] = 1;

    for(int i = head[u]; i != -1;i = edge[i].next )

    {

        int v = edge[i].to;

        if(v == pre)

            continue;

        dfs(v,len+1,u);

        Size[u] += Size[v];

    }

}

void solve(int u,ll tans,int pre)

{

    ans[u] = tans;

    for(int i = head[u];i != -1;i = edge[i].next)

    {

        int v = edge[i].to;

        if(v == pre)

            continue;

        ll t = n-Size[v]*2;

        solve(v, (ll)tans + t,u);

    }

}

void init()

{

    tot = 0;

    clr(head,-1);

}

int main()

{

    int u,v;

    while(scanfi(n) != EOF)

    {

        init();

        for(int i = 1;i < n;i++)

        {

            scanfi(u),scanfi(v);

            add_edge(u,v);

            add_edge(v,u);

        }

        dfs(1,0,-1);

        ll ta = 0;

        for(int i =1;i <= n;i++)

          ta += f[i];

        solve(1,ta,-1);

        for(int i = 1;i <= n;i++)

        {

            print(ans[i]);

        }

    }

}