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;
namespace FGF {
const int N = 5e5 + 5;
struct edg {
  int to, nxt, w;
} e[N << 1];
int head[N], cnt, dis[N], L[N], R[N], num, id[N], ans[N], dep[N], n, son[N],
    siz[N];
int mx[(1 << 22) + 5];
void add(int u, int v, int w) {
  cnt++;
  e[cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
  e[cnt].w = w;
}
void dfs1(int u, int f) {
  siz[u] = 1;
  L[u] = ++num, id[num] = u, dep[u] = dep[f] + 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    dis[v] = dis[u] ^ e[i].w, dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
  R[u] = num;
}
void dfs2(int u, int op) {
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].to != son[u]) dfs2(e[i].to, 0), ans[u] = max(ans[u], ans[e[i].to]);
  if (son[u]) dfs2(son[u], 1), ans[u] = max(ans[u], ans[son[u]]);
  if (mx[dis[u]]) ans[u] = max(ans[u], mx[dis[u]] - dep[u]);
  for (int k = 0; k < 22; k++)
    if (mx[dis[u] ^ (1 << k)])
      ans[u] = max(ans[u], mx[dis[u] ^ (1 << k)] - dep[u]);
  mx[dis[u]] = max(mx[dis[u]], dep[u]);
  for (int i = head[u]; i; i = e[i].nxt)
    if (e[i].to != son[u]) {
      int v = e[i].to;
      for (int j = L[v]; j <= R[v]; j++) {
        if (mx[dis[id[j]]])
          ans[u] = max(ans[u], mx[dis[id[j]]] + dep[id[j]] - 2 * dep[u]);
        for (int k = 0; k < 22; k++)
          if (mx[dis[id[j]] ^ (1 << k)])
            ans[u] = max(ans[u],
                         mx[dis[id[j]] ^ (1 << k)] + dep[id[j]] - 2 * dep[u]);
      }
      for (int j = L[v]; j <= R[v]; j++)
        mx[dis[id[j]]] = max(mx[dis[id[j]]], dep[id[j]]);
    }
  if (!op)
    for (int i = L[u]; i <= R[u]; i++) mx[dis[id[i]]] = 0;
}
void work() {
  scanf("%d", &n);
  char ch[5];
  int x;
  for (int i = 2; i <= n; i++)
    scanf("%d%s", &x, ch), add(x, i, 1 << (ch[0] - 'a'));
  dfs1(1, 0), dfs2(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
}  // namespace FGF
int main() {
  FGF::work();
  return 0;
}
