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 = 5e5 + 10;
const int INF = 0x3f3f3f3f;
int n;
int h[N], e[N], ne[N], w[N], idx;
int L[N], R[N], tot, id[N];
int f[N * 10], d[N], size[N], son[N], Son, dis[N];
int ans[N];
void add(int a, int b, int c) {
  w[idx] = c;
  e[idx] = b;
  ne[idx] = h[a];
  h[a] = idx++;
}
void dfs(int u, int fa) {
  size[u] = 1;
  L[u] = ++tot;
  id[tot] = u;
  d[u] = d[fa] + 1;
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i];
    if (j == fa) continue;
    dis[j] = dis[u] ^ w[i];
    dfs(j, u);
    size[u] += size[j];
    if (size[son[u]] < size[j]) son[u] = j;
  }
  R[u] = tot;
}
void calc(int u, int fa) {
  if (f[dis[u]]) ans[u] = max(ans[u], f[dis[u]] - d[u]);
  for (int i = 0; i < 22; i++) {
    if (f[dis[u] ^ (1 << i)]) {
      ans[u] = max(ans[u], f[dis[u] ^ (1 << i)] - d[u]);
    }
  }
  f[dis[u]] = max(f[dis[u]], d[u]);
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i];
    if (j == fa || j == Son) continue;
    for (int k = L[j]; k <= R[j]; k++) {
      int x = id[k];
      if (f[dis[x]]) ans[u] = max(ans[u], f[dis[x]] + d[x] - 2 * d[u]);
      for (int z = 0; z < 22; z++) {
        if (f[dis[x] ^ (1 << z)]) {
          ans[u] = max(ans[u], f[dis[x] ^ (1 << z)] + d[x] - 2 * d[u]);
        }
      }
    }
    for (int k = L[j]; k <= R[j]; k++) {
      int x = id[k];
      f[dis[x]] = max(f[dis[x]], d[x]);
    }
  }
}
void dsu(int u, int fa, int opt) {
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i];
    if (j == fa || j == son[u]) continue;
    dsu(j, u, 0);
    ans[u] = max(ans[u], ans[j]);
  }
  if (son[u])
    dsu(son[u], u, 1), Son = son[u], ans[u] = max(ans[u], ans[son[u]]);
  calc(u, fa);
  Son = 0;
  if (!opt) {
    for (int i = L[u]; i <= R[u]; i++) f[dis[id[i]]] = 0;
  }
}
int main() {
  memset(h, -1, sizeof(h));
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    char s[10];
    scanf("%d%s", &x, s);
    int t = s[0] - 'a';
    add(x, i, 1ll << t);
  }
  dfs(1, 0);
  dsu(1, 0, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  puts("");
}
