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.

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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;
const int N = 500010;
int Xor[N], son[N], n;
struct node {
  int to, nxt;
} g[N];
int lst[N], gl, w[N], siz[N], dep[N], L[N], R[N], tot, dfn[N];
void add(int x, int y) {
  g[++gl] = (node){y, lst[x]};
  lst[x] = gl;
}
void dfs1(int u) {
  siz[u] = 1;
  dfn[++tot] = u;
  L[u] = tot;
  for (int i = lst[u]; i; i = g[i].nxt) {
    int v = g[i].to;
    Xor[v] = Xor[u] ^ w[v];
    dep[v] = dep[u] + 1;
    dfs1(v);
    siz[u] += siz[v];
    if (!son[u] || siz[v] > siz[son[u]]) son[u] = v;
  }
  R[u] = tot;
}
int f[1 << 22], ans[N];
void dfs2(int u, bool op) {
  for (int i = lst[u]; i; i = g[i].nxt)
    if (g[i].to != son[u]) dfs2(g[i].to, 0), ans[u] = max(ans[u], ans[g[i].to]);
  if (son[u]) dfs2(son[u], 1), ans[u] = max(ans[u], ans[son[u]]);
  if (f[Xor[u]]) ans[u] = max(ans[u], f[Xor[u]] - dep[u]);
  for (int i = 0; i < 22; i++)
    if (f[Xor[u] ^ (1 << i)])
      ans[u] = max(ans[u], f[Xor[u] ^ (1 << i)] - dep[u]);
  f[Xor[u]] = max(f[Xor[u]], dep[u]);
  for (int i = lst[u]; i; i = g[i].nxt) {
    int v = g[i].to;
    if (v != son[u]) {
      for (int i = L[v]; i <= R[v]; i++) {
        if (f[Xor[dfn[i]]])
          ans[u] = max(ans[u], f[Xor[dfn[i]]] + dep[dfn[i]] - 2 * dep[u]);
        for (int j = 0; j < 22; j++)
          if (f[Xor[dfn[i]] ^ (1 << j)])
            ans[u] = max(ans[u],
                         f[Xor[dfn[i]] ^ (1 << j)] + dep[dfn[i]] - 2 * dep[u]);
      }
      for (int i = L[v]; i <= R[v]; i++)
        f[Xor[dfn[i]]] = max(f[Xor[dfn[i]]], dep[dfn[i]]);
    }
  }
  if (!op)
    for (int i = L[u]; i <= R[u]; i++) f[Xor[dfn[i]]] = 0;
}
int main() {
  scanf("%d", &n);
  char c;
  for (int i = 2, f; i <= n; i++) {
    cin >> f >> c;
    add(f, i);
    w[i] = 1 << (c - 'a');
  }
  dfs1(1);
  dfs2(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
