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;
int u, n, fa[500010], faw[500010];
int w[500010], d[500010], sz[500010], ans[500010], son[500010], now;
int cnt[(1 << 22)];
char c;
vector<int> a[500010];
void prepare(int x) {
  sz[x] = 1;
  for (int i = 0; i < a[x].size(); i++) {
    int u = a[x][i];
    w[u] = w[x] + 1;
    d[u] = d[u] ^ d[x];
    prepare(u);
    sz[x] += sz[u];
    if (sz[u] > sz[son[x]]) son[x] = u;
  }
}
void add(int x) {
  cnt[d[x]] = max(cnt[d[x]], w[x]);
  for (int i = 0; i < a[x].size(); i++) add(a[x][i]);
}
void check(int x) {
  ans[now] = max(ans[now], w[x] + cnt[d[x]]);
  for (int i = 0; i < 22; i++)
    ans[now] = max(ans[now], w[x] + cnt[(1 << i) ^ d[x]]);
  for (int i = 0; i < a[x].size(); i++) check(a[x][i]);
}
void del(int x) {
  cnt[d[x]] = -0x3FFFFFFF;
  for (int i = 0; i < a[x].size(); i++) del(a[x][i]);
}
void dfs(int x) {
  for (int i = 0; i < a[x].size(); i++)
    if (a[x][i] != son[x]) {
      dfs(a[x][i]);
      del(a[x][i]);
    }
  if (son[x]) dfs(son[x]);
  now = x;
  for (int i = 0; i < a[x].size(); i++)
    if (a[x][i] != son[x]) {
      check(a[x][i]);
      add(a[x][i]);
    }
  cnt[d[x]] = max(cnt[d[x]], w[x]);
  ans[x] = max(ans[x], w[x] + cnt[d[x]]);
  for (int i = 0; i < 22; i++)
    ans[x] = max(ans[x], w[x] + cnt[(1 << i) ^ d[x]]);
  ans[x] -= 2 * w[x];
  for (int i = 0; i < a[x].size(); i++) ans[x] = max(ans[x], ans[a[x][i]]);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (1 << 22); i++) cnt[i] = -0x3FFFFFFF;
  for (int i = 2; i <= n; i++) {
    scanf("%d %c", &u, &c);
    a[u].push_back(i);
    fa[i] = u;
    d[i] = 1 << (c - 'a');
  }
  prepare(1);
  dfs(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
