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 MAX_N = 5e5 + 200;
int head[MAX_N], current, n, fa[MAX_N], siz[MAX_N], lft[MAX_N], rig[MAX_N],
    ptot;
int son[MAX_N], dep[MAX_N], anti[MAX_N], book[1 << 23], answer[MAX_N],
    dist[MAX_N];
char opt[10];
struct edge {
  int to, nxt, weight;
} edges[MAX_N << 1];
void addpath(int src, int dst, int weight) {
  edges[current].to = dst, edges[current].nxt = head[src];
  edges[current].weight = weight, head[src] = current++;
}
void predfs(int u) {
  lft[u] = ++ptot, siz[u] = 1, dep[u] = dep[fa[u]] + 1, anti[ptot] = u;
  for (int i = head[u]; i != -1; i = edges[i].nxt) {
    dist[edges[i].to] = dist[u] ^ edges[i].weight, predfs(edges[i].to),
    siz[u] += siz[edges[i].to];
    if (siz[son[u]] < siz[edges[i].to]) son[u] = edges[i].to;
  }
  rig[u] = ptot;
}
void dfs(int u, bool save) {
  for (int i = head[u]; i != -1; i = edges[i].nxt)
    if (edges[i].to != son[u])
      dfs(edges[i].to, false), answer[u] = max(answer[u], answer[edges[i].to]);
  if (son[u] != 0)
    dfs(son[u], true), answer[u] = max(answer[u], answer[son[u]]);
  if (book[dist[u]] != 0) answer[u] = max(answer[u], book[dist[u]] - dep[u]);
  for (int ch = 0; ch <= 21; ch++)
    if (book[dist[u] ^ (1 << ch)])
      answer[u] = max(answer[u], book[dist[u] ^ (1 << ch)] - dep[u]);
  book[dist[u]] = max(book[dist[u]], dep[u]);
  for (int i = head[u]; i != -1; i = edges[i].nxt)
    if (edges[i].to != son[u]) {
      for (int id = lft[edges[i].to]; id <= rig[edges[i].to]; id++) {
        int curt = anti[id];
        if (book[dist[curt]])
          answer[u] =
              max(answer[u], dep[curt] + book[dist[curt]] - (dep[u] << 1));
        for (int ch = 0; ch <= 21; ch++)
          if (book[dist[curt] ^ (1 << ch)])
            answer[u] = max(answer[u], book[dist[curt] ^ (1 << ch)] +
                                           dep[curt] - (dep[u] << 1));
      }
      for (int id = lft[edges[i].to]; id <= rig[edges[i].to]; id++)
        book[dist[anti[id]]] = max(book[dist[anti[id]]], dep[anti[id]]);
    }
  if (save == false)
    for (int id = lft[u]; id <= rig[u]; id++) book[dist[anti[id]]] = 0;
}
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = 2; i <= n; i++)
    scanf("%d%s", &fa[i], opt + 1), addpath(fa[i], i, 1 << (opt[1] - 'a'));
  predfs(1), dfs(1, true);
  for (int i = 1; i <= n; i++) printf("%d ", answer[i]);
  return 0;
}
