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;
vector<int> gph[500005];
int n, par[500005], dep[500005], msk[500005];
int dp[500005], sz[500005];
int mx[1 << 22];
int l[23] = {};
void dfs2(int x, vector<pair<int, int> > &v) {
  v.push_back(pair<int, int>(dep[x], msk[x]));
  for (auto &i : gph[x]) dfs2(i, v);
}
void dfs(int x) {
  if (gph[x].empty()) {
    mx[msk[x]] = dep[x];
    return;
  }
  for (int i = 1; i < gph[x].size(); i++) {
    dfs(gph[x][i]);
    vector<pair<int, int> > w;
    dfs2(gph[x][i], w);
    for (auto &j : w) mx[j.second] = -1;
  }
  dfs(gph[x][0]);
  for (int i = 1; i < gph[x].size(); i++) {
    vector<pair<int, int> > w;
    dfs2(gph[x][i], w);
    for (auto &j : w) {
      for (int k = 0; k < 23; k++) {
        int val = mx[j.second ^ l[k]];
        if (~val) dp[x] = max(dp[x], j.first + val - 2 * dep[x]);
      }
    }
    for (auto &j : w) mx[j.second] = max(mx[j.second], j.first);
  }
  for (int k = 0; k < 23; k++) {
    if (mx[msk[x] ^ l[k]] != -1) {
      dp[x] = max(dp[x], mx[msk[x] ^ l[k]] - dep[x]);
    }
  }
  mx[msk[x]] = max(mx[msk[x]], dep[x]);
}
int main() {
  scanf("%d", &n);
  char ch[6];
  for (int i = 0; i < 22; i++) l[i] = (1 << i);
  for (int i = 2; i <= n; i++) {
    scanf("%d %s", &par[i], ch);
    gph[par[i]].push_back(i);
    dep[i] = dep[par[i]] + 1;
    msk[i] = msk[par[i]] ^ (1 << (*ch - 'a'));
  }
  for (int i = n; i; i--) {
    sz[i] = 1;
    for (auto &j : gph[i]) sz[i] += sz[j];
    sort(gph[i].begin(), gph[i].end(),
         [&](const int &a, const int &b) { return sz[a] > sz[b]; });
  }
  memset(mx, -1, sizeof(mx));
  dfs(1);
  for (int i = n; i; i--) {
    for (auto &j : gph[i]) dp[i] = max(dp[i], dp[j]);
  }
  for (int i = 1; i <= n; i++) printf("%d ", dp[i]);
}
