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;
inline int read() {
  int ans = 0, fh = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  return ans * fh;
}
const int maxn = 1e6 + 100;
int n, head[maxn], nex[maxn << 1], v[maxn << 1], num = 1;
int md[maxn], dep[maxn], son[maxn], ans[maxn];
int *f[maxn], tmp[maxn], *id = tmp;
inline void add(int x, int y) {
  v[++num] = y, nex[num] = head[x];
  head[x] = num;
  v[++num] = x, nex[num] = head[y];
  head[y] = num;
}
void dfs1(int x, int Fa) {
  md[x] = dep[x] = dep[Fa] + 1;
  for (int i = head[x]; i; i = nex[i]) {
    int y = v[i];
    if (y == Fa) continue;
    dfs1(y, x), (md[x] = max(md[x], md[y]));
    if (md[y] > md[son[x]]) son[x] = y;
  }
}
void dfs2(int x, int Fa) {
  if (son[x]) {
    f[son[x]] = f[x] + 1, dfs2(son[x], x);
    ans[x] = ans[son[x]] + 1;
  }
  f[x][0] = 1;
  for (int i = head[x]; i; i = nex[i]) {
    int y = v[i];
    if (y == Fa || y == son[x]) continue;
    f[y] = id, id += md[y] - dep[y] + 1, dfs2(y, x);
    for (int j = 0; j <= md[y] - dep[y]; j++) {
      f[x][j + 1] += f[y][j];
      if (f[x][j + 1] > f[x][ans[x]] ||
          (f[x][j + 1] == f[x][ans[x]] && j + 1 < ans[x]))
        ans[x] = j + 1;
    }
  }
  if (f[x][ans[x]] == 1) ans[x] = 0;
}
int main() {
  n = read();
  for (int i = 1; i < n; i++) add(read(), read());
  dfs1(1, 0), f[1] = id, id += md[1], dfs2(1, 0);
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
