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 n, ans[500005], son[500005], nx[(1 << 22) + 5], dis[500005], sz[500005],
    tot, w[500005], id[500005], dep[500005];
char x[5];
struct node {
  int x, y;
};
vector<node> v[500005];
void dfs(int x) {
  sz[x] = 1;
  w[++tot] = x;
  id[x] = tot;
  for (int i = 0; i < v[x].size(); i++) {
    int h = v[x][i].x, y = v[x][i].y;
    dis[h] = dis[x] ^ y;
    dep[h] = dep[x] + 1;
    dfs(h);
    sz[x] += sz[h];
    if (sz[h] > sz[son[x]]) son[x] = h;
  }
}
void dsu(int x, int fl) {
  for (int i = 0; i < v[x].size(); i++) {
    int h = v[x][i].x;
    if (son[x] == h) continue;
    dsu(h, 0);
    ans[x] = max(ans[x], ans[h]);
  }
  if (son[x]) {
    dsu(son[x], 1);
    ans[x] = max(ans[x], ans[son[x]]);
  }
  if (nx[dis[x]]) ans[x] = max(ans[x], nx[dis[x]] - dep[x]);
  for (int i = 0; i < 22; i++)
    if (nx[dis[x] ^ (1 << i)])
      ans[x] = max(ans[x], nx[dis[x] ^ (1 << i)] - dep[x]);
  nx[dis[x]] = max(nx[dis[x]], dep[x]);
  for (int i = 0; i < v[x].size(); i++) {
    int h = v[x][i].x;
    if (h == son[x]) continue;
    for (int j = id[h]; j < id[h] + sz[h]; j++) {
      int u = w[j];
      if (nx[dis[u]]) ans[x] = max(ans[x], nx[dis[u]] + dep[u] - 2 * dep[x]);
      for (int k = 0; k < 22; k++)
        if (nx[dis[u] ^ (1 << k)])
          ans[x] = max(ans[x], nx[dis[u] ^ (1 << k)] + dep[u] - 2 * dep[x]);
    }
    for (int j = id[h]; j < id[h] + sz[h]; j++) {
      int u = w[j];
      nx[dis[u]] = max(nx[dis[u]], dep[u]);
    }
  }
  if (!fl)
    for (int i = id[x]; i < id[x] + sz[x]; i++) nx[dis[w[i]]] = 0;
}
signed main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int y;
    cin >> y >> x;
    v[y].push_back((node){i, 1 << (x[0] - 'a')});
  }
  dep[1] = 1;
  dfs(1);
  dsu(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
