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 maxn = 5e5 + 1;
int in[maxn], out[maxn], v[maxn], n, sz[maxn], t, id[maxn], son[maxn],
    ans[maxn], h[maxn];
vector<int> g[maxn];
void sz_dfs(int cur) {
  sz[cur] = 1;
  in[cur] = ++t;
  id[t] = cur;
  for (auto nxt : g[cur]) {
    v[nxt] ^= v[cur];
    h[nxt] = h[cur] + 1;
    sz_dfs(nxt);
    sz[cur] += sz[nxt];
    if (sz[nxt] > sz[son[cur]]) son[cur] = nxt;
  }
  out[cur] = t;
}
const int B = 22;
const int maxb = 1 << B;
int cnt[maxb];
inline void Max(int &a, int b) {
  if (a < b) a = b;
}
void dfs(int cur, bool keep) {
  for (auto nxt : g[cur])
    if (nxt != son[cur]) {
      dfs(nxt, 0);
      Max(ans[cur], ans[nxt]);
    }
  if (son[cur]) {
    dfs(son[cur], true);
    Max(ans[cur], ans[son[cur]]);
  }
  for (auto nxt : g[cur])
    if (nxt != son[cur]) {
      for (int i = in[nxt]; i <= out[nxt]; ++i) {
        int dh = h[id[i]] - h[cur];
        int tar = v[id[i]];
        for (int i = 0; i < B; ++i) {
          int tmp = tar ^ (1 << i);
          if (cnt[tmp]) Max(ans[cur], dh + cnt[tmp] - h[cur]);
        }
        if (cnt[tar]) Max(ans[cur], dh + cnt[tar] - h[cur]);
      }
      for (int i = in[nxt]; i <= out[nxt]; ++i) Max(cnt[v[id[i]]], h[id[i]]);
    }
  int tar = v[cur];
  for (int i = 0; i < B; ++i) {
    int tmp = tar ^ (1 << i);
    if (cnt[tmp]) Max(ans[cur], cnt[tmp] - h[cur]);
  }
  if (cnt[tar]) Max(ans[cur], cnt[tar] - h[cur]);
  Max(cnt[v[cur]], h[cur]);
  if (!keep)
    for (int i = in[cur]; i <= out[cur]; ++i) cnt[v[id[i]]] = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int p;
    char c;
    scanf("%d %c", &p, &c);
    c -= 'a';
    v[i] = 1 << c;
    g[p].push_back(i);
  }
  sz_dfs(1);
  dfs(1, 1);
  for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], i == n ? '\n' : ' ');
}
