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 + 5;
const int M = 5e6 + 5;
struct Edge {
  int to, nxt;
} E[N];
int n, c[N], cnt, head[N], f[M], t, q[N], s[N];
int son[N], ans[N], l[N], st[N], depth[N];
void addedg(int u, int v) {
  E[++cnt].to = v;
  E[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs1(int u) {
  s[u] = 1;
  for (int i = head[u]; i; i = E[i].nxt) {
    int v = E[i].to;
    depth[v] = depth[u] + 1;
    st[v] = st[u] ^ (1 << c[v]);
    dfs1(v);
    s[u] += s[v];
    if (s[v] > s[son[u]]) son[u] = v;
  }
}
void remove(int u) {
  f[st[u]] = 0;
  for (int i = head[u]; i; i = E[i].nxt) remove(E[i].to);
}
void calc(int k, int u) {
  int now = st[k] ^ st[u];
  if ((now & -now) == now || !now) ans[k] = max(ans[k], depth[u] - depth[k]);
  if (f[st[u]]) ans[k] = max(ans[k], f[st[u]] + depth[u] - 2 * depth[k]);
  for (int i = 0; i < 22; i++)
    if (f[st[u] ^ (1 << i)])
      ans[k] = max(ans[k], f[st[u] ^ (1 << i)] + depth[u] - 2 * depth[k]);
  for (int i = head[u]; i; i = E[i].nxt) {
    int v = E[i].to;
    calc(k, v);
  }
}
void add(int u) {
  f[st[u]] = max(f[st[u]], depth[u]);
  for (int i = head[u]; i; i = E[i].nxt) {
    int v = E[i].to;
    add(v);
  }
}
void dfs2(int u) {
  if (!son[u]) return;
  for (int i = head[u]; i; i = E[i].nxt) {
    int v = E[i].to;
    if (v == son[u]) continue;
    dfs2(v);
    remove(v);
    ans[u] = max(ans[u], ans[v]);
  }
  dfs2(son[u]);
  f[st[son[u]]] = max(depth[son[u]], f[st[son[u]]]);
  ans[u] = max(ans[u], max(ans[son[u]], f[st[u]] - depth[u]));
  for (int i = 0; i < 22; i++)
    ans[u] = max(ans[u], f[st[u] ^ (1 << i)] - depth[u]);
  for (int i = head[u]; i; i = E[i].nxt) {
    int v = E[i].to;
    if (v == son[u]) continue;
    calc(u, v);
    add(v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    char opt[3];
    scanf("%d", &p);
    scanf("%s", opt);
    addedg(p, i);
    c[i] = opt[0] - 'a';
  }
  dfs1(1);
  dfs2(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
