Just in case somebody missed it: we have wonderful girls in Arpa’s land.

Arpa has a rooted tree (connected acyclic graph) consisting of n vertices. The vertices are numbered 1 through n, the vertex 1 is the root. There is a letter written on each edge of this tree. Mehrdad is a fan of Dokhtar-kosh things. He call a string Dokhtar-kosh, if we can shuffle the characters in string such that it becomes palindrome.

<image>

He asks Arpa, for each vertex v, what is the length of the longest simple path in subtree of v that form a Dokhtar-kosh string.

Input

The first line contains integer n (1 ≤ n ≤ 5·105) — the number of vertices in the tree.

(n - 1) lines follow, the i-th of them contain an integer pi + 1 and a letter ci + 1 (1 ≤ pi + 1 ≤ i, ci + 1 is lowercase English letter, between a and v, inclusively), that mean that there is an edge between nodes pi + 1 and i + 1 and there is a letter ci + 1 written on this edge.

Output

Print n integers. The i-th of them should be the length of the longest simple path in subtree of the i-th vertex that form a Dokhtar-kosh string.

Examples

Input

4
1 s
2 a
3 s


Output

3 1 1 0 

Input

5
1 a
2 h
1 a
4 h


Output

4 1 0 1 0 
#include <bits/stdc++.h>
using namespace std;
int n, ci[1 << 22], son[500010], siz[500010], dep[500010];
int hd[500010], nt[500010], D[500010], ans[500010];
int getint() {
  int w = 0;
  bool q = 0;
  char c = getchar();
  while ((c > '9' || c < '0') && c != '-') c = getchar();
  if (c == '-') c = getchar(), q = 1;
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? -w : w;
}
void dfs(int u) {
  siz[u] = 1;
  for (int i = hd[u]; i; i = nt[i]) {
    D[i] ^= D[u];
    dep[i] = dep[u] + 1;
    dfs(i);
    siz[u] += siz[i];
    if (siz[i] > siz[son[u]]) son[u] = i;
  }
}
void undo(int u) {
  ci[D[u]] = -(1 << 30);
  for (int i = hd[u]; i; i = nt[i]) undo(i);
}
int o;
void up(int& x, int y) {
  if (y > x) x = y;
}
void up(int u) {
  up(ans[o], dep[u] + ci[D[u]]);
  for (int i = 0; i <= 21; i++) up(ans[o], dep[u] + ci[1 << i ^ D[u]]);
  for (int i = hd[u]; i; i = nt[i]) up(i);
}
void ins(int u) {
  up(ci[D[u]], dep[u]);
  for (int i = hd[u]; i; i = nt[i]) ins(i);
}
void work(int u) {
  for (int i = hd[u]; i; i = nt[i])
    if (i != son[u]) work(i), undo(i);
  if (son[u]) work(son[u]);
  o = u;
  for (int i = hd[u]; i; i = nt[i])
    if (i != son[u]) up(i), ins(i);
  up(ci[D[u]], dep[u]);
  up(ans[u], dep[u] + ci[D[u]]);
  for (int i = 0; i <= 21; i++) up(ans[u], dep[u] + ci[1 << i ^ D[u]]);
  ans[u] -= dep[u] << 1;
  for (int i = hd[u]; i; i = nt[i]) up(ans[u], ans[i]);
}
int main() {
  n = getint();
  for (int i = 0; i < (1 << 22); i++) ci[i] = -(1 << 30);
  for (int i = 2, x; i <= n; i++) {
    x = getint();
    nt[i] = hd[x];
    hd[x] = i;
    char c = getchar();
    while (c > 'v' || c < 'a') c = getchar();
    D[i] = 1 << (c - 'a');
  }
  dfs(1);
  work(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
