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>
const double eps = 1e-8;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 5e5 + 7;
using namespace std;
struct node {
  int v, w;
};
vector<node> edge[maxn];
int siz[maxn], hson[maxn], dis[maxn];
int ans[maxn], dep[maxn], f[maxn * 20];
void dfs1(int u, int fa) {
  siz[u] = 1;
  dep[u] = dep[fa] + 1;
  for (auto e : edge[u]) {
    if (e.v == fa) continue;
    dis[e.v] ^= e.w ^ dis[u];
    dfs1(e.v, u);
    siz[u] += siz[e.v];
    if (siz[hson[u]] < siz[e.v]) hson[u] = e.v;
  }
}
int flag = 0;
void calc(int u, int fa, int root) {
  ans[root] = max(ans[root], f[dis[u]] + dep[u] - 2 * dep[root]);
  if ((dis[u] ^ dis[root]) == 0) ans[root] = max(ans[root], dep[u] - dep[root]);
  for (int i = 0; i < 22; i++) {
    int now = dis[u] ^ (1 << i);
    ans[root] = max(ans[root], f[now] + dep[u] - 2 * dep[root]);
    if ((now ^ dis[root]) == 0) ans[root] = max(ans[root], dep[u] - dep[root]);
  }
  for (auto e : edge[u]) {
    calc(e.v, u, root);
  }
}
void update(int u, int fa, int op) {
  if (op == 1) {
    f[dis[u]] = max(f[dis[u]], dep[u]);
  } else {
    f[dis[u]] = -inf;
  }
  for (auto e : edge[u]) update(e.v, u, op);
}
void dfs2(int u, int fa, int keep) {
  for (auto e : edge[u]) {
    if (e.v == hson[u]) continue;
    dfs2(e.v, u, 0);
  }
  if (hson[u]) {
    dfs2(hson[u], u, 1);
  }
  ans[u] = max(ans[u], f[dis[u]] - dep[u]);
  for (int i = 0; i < 22; i++) {
    int now = dis[u] ^ (1 << i);
    ans[u] = max(ans[u], f[now] - dep[u]);
  }
  for (auto e : edge[u]) {
    if (e.v == hson[u]) continue;
    calc(e.v, u, u);
    update(e.v, u, 1);
  }
  if (!keep) {
    for (auto e : edge[u]) update(e.v, u, -1);
    f[dis[u]] = -inf;
  } else
    f[dis[u]] = max(f[dis[u]], dep[u]);
}
void dfs3(int u, int fa) {
  for (auto e : edge[u]) {
    if (e.v == fa) continue;
    dfs3(e.v, u);
    ans[u] = max(ans[u], ans[e.v]);
  }
}
int main(void) {
  fill(f, f + 20 * maxn, -inf);
  int n;
  scanf("%d", &n);
  for (int u = 2; u <= n; u++) {
    int v;
    char c;
    scanf("%d %c", &v, &c);
    edge[v].push_back(node{u, 1 << (c - 'a')});
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  dfs3(1, 0);
  for (int i = 1; i <= n; i++) printf("%d%c", ans[i], " \n"[i == n]);
  return 0;
}
