You are given a rooted undirected tree consisting of n vertices. Vertex 1 is the root.

Let's denote a depth array of vertex x as an infinite sequence [d_{x, 0}, d_{x, 1}, d_{x, 2}, ...], where d_{x, i} is the number of vertices y such that both conditions hold:

  * x is an ancestor of y; 
  * the simple path from x to y traverses exactly i edges. 



The dominant index of a depth array of vertex x (or, shortly, the dominant index of vertex x) is an index j such that:

  * for every k < j, d_{x, k} < d_{x, j}; 
  * for every k > j, d_{x, k} ≤ d_{x, j}. 



For every vertex in the tree calculate its dominant index.

Input

The first line contains one integer n (1 ≤ n ≤ 10^6) — the number of vertices in a tree.

Then n - 1 lines follow, each containing two integers x and y (1 ≤ x, y ≤ n, x ≠ y). This line denotes an edge of the tree.

It is guaranteed that these edges form a tree.

Output

Output n numbers. i-th number should be equal to the dominant index of vertex i.

Examples

Input

4
1 2
2 3
3 4


Output

0
0
0
0


Input

4
1 2
1 3
1 4


Output

1
0
0
0


Input

4
1 2
2 3
2 4


Output

2
1
0
0
#include <bits/stdc++.h>
using namespace std;
vector<int> g[2002000], vec[2002000];
int dep[2002000], last[2002000], mx[2002000], ans[2002000], top[2002000];
int n, m;
void dfs1(int x, int fa) {
  last[x] = dep[x] = dep[fa] + 1;
  for (int i = 0, rr = (int)g[x].size(); i < rr; i++) {
    int too = g[x][i];
    if (too == fa) continue;
    dfs1(too, x);
    if (last[too] > last[mx[x]]) mx[x] = too;
  }
  if (g[x].size() > 1 || !fa) last[x] = last[mx[x]];
}
void dfs2(int x, int fa, int anc) {
  if (mx[x]) {
    dfs2(mx[x], x, anc);
    ans[x] = ans[mx[x]];
  }
  for (auto &too : g[x]) {
    if (too == fa || too == mx[x]) continue;
    dfs2(too, x, too);
    int l = last[mx[x]] - last[too], r = last[too] - dep[too];
    for (int i = 0; i <= r; i++) {
      vec[anc][i + l] += vec[too][i];
      if (vec[anc][i + l] > vec[anc][ans[x]] ||
          (vec[anc][i + l] == vec[anc][ans[x]] && ans[x] < i + l))
        ans[x] = i + l;
    }
  }
  vec[anc].push_back(1);
  if (vec[anc][ans[x]] == 1) ans[x] = last[x] - dep[x];
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int j, k;
    scanf("%d %d", &j, &k);
    g[j].push_back(k);
    g[k].push_back(j);
  }
  dep[0] = -1;
  dfs1(1, 0);
  dfs2(1, 0, 1);
  for (int i = 1; i <= n; i++) printf("%d\n", last[i] - dep[i] - ans[i]);
}
int main() {
  solve();
  return 0;
}
