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;
const int N = 1e6 + 50;
inline int rd(register int x = 0, register char ch = getchar(),
              register int f = 0) {
  for (; !isdigit(ch); ch = getchar()) f = ch == '-';
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - 48;
  return f ? -x : x;
}
int n, B;
int head[N], to[N << 1], nxt[N << 1], tmp[N], *id = tmp, *f[N], len[N], hson[N],
                                              ans[N];
void lnk(int x, int y) {
  to[++B] = y, nxt[B] = head[x], head[x] = B;
  to[++B] = x, nxt[B] = head[y], head[y] = B;
}
void dfs1(int x, int prt) {
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != prt)
      dfs1(to[i], x), (len[hson[x]] > len[to[i]] ? 0 : hson[x] = to[i]);
  len[x] = len[hson[x]] + 1;
}
void dfs2(int x, int prt) {
  f[x][0] = 1;
  if (hson[x])
    f[hson[x]] = f[x] + 1, dfs2(hson[x], x), ans[x] = ans[hson[x]] + 1;
  for (int i = head[x]; i; i = nxt[i])
    if (to[i] != prt && to[i] != hson[x]) {
      f[to[i]] = id, id += len[to[i]], dfs2(to[i], x);
      for (int j = 1; j <= len[to[i]]; ++j) {
        f[x][j] += f[to[i]][j - 1];
        if (f[x][j] > f[x][ans[x]] || (f[x][j] == f[x][ans[x]] && j < ans[x]))
          ans[x] = j;
      }
    }
  if (f[x][ans[x]] == 1) ans[x] = 0;
}
int main() {
  n = rd();
  for (int i = 1; i < n; ++i) lnk(rd(), rd());
  dfs1(1, 0), f[1] = id, id += len[1], dfs2(1, 0);
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
}
