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;
const int N = 5e5 + 10;
struct Edge {
  int to, nxt, val;
} e[N << 1];
int h[N], idx;
void Ins(int a, int b, int c) {
  e[++idx].to = b;
  e[idx].nxt = h[a];
  h[a] = idx;
  e[idx].val = c;
}
int ans[N], f[1 << 22], dep[N], d[N], siz[N], son[N];
void dfs1(int u) {
  siz[u] = 1;
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    dep[v] = dep[u] + 1;
    d[v] = d[u] ^ (1 << e[i].val);
    dfs1(v);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
int Max;
void clear(int u) {
  f[d[u]] = 0;
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    clear(v);
  }
}
int now;
void calc(int u) {
  if (f[d[u]]) Max = max(Max, dep[u] + f[d[u]] - now);
  for (int i = 0; i < 22; i++) {
    int x = f[d[u] ^ (1 << i)];
    if (x) Max = max(Max, dep[u] + x - now);
  }
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    calc(v);
  }
}
void update(int u) {
  f[d[u]] = max(f[d[u]], dep[u]);
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    update(v);
  }
}
void dfs2(int u) {
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == son[u]) continue;
    dfs2(v);
    clear(v);
    Max = 0;
  }
  if (son[u]) dfs2(son[u]);
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    Max = max(Max, ans[v]);
  }
  now = dep[u] << 1;
  for (int i = h[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == son[u]) continue;
    calc(v);
    update(v);
  }
  if (f[d[u]]) Max = max(Max, dep[u] + f[d[u]] - now);
  for (int i = 0; i < 22; i++) {
    int x = f[d[u] ^ (1 << i)];
    if (x) Max = max(Max, dep[u] + x - now);
  }
  f[d[u]] = max(f[d[u]], dep[u]);
  ans[u] = Max;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    char s[10];
    int x;
    scanf("%d%s", &x, s);
    Ins(x, i, s[0] - 'a');
  }
  dfs1(1);
  dfs2(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
