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;
const int N = 5e5 + 10, S = 1 << 23;
int tot, cnt, n, dfn[N], out[N], f[S], rk[N], ans[N], sum[N], dep[N], head[N],
    to[N], ne[N], w[N], size[N], son[N], g[N][23];
char t[11];
void add(int x, int y, int z) {
  to[++tot] = y;
  w[tot] = z;
  ne[tot] = head[x];
  head[x] = tot;
}
void dfs1(int x) {
  dfn[x] = ++cnt;
  rk[cnt] = x;
  size[x] = 1;
  for (int i = head[x]; i; i = ne[i]) {
    int y = to[i];
    dep[y] = dep[x] + 1;
    sum[y] = sum[x] ^ (1 << w[i]);
    dfs1(y);
    if (size[y] > size[son[x]]) son[x] = y;
    size[x] += size[y];
  }
  g[x][22] = sum[x];
  for (int j = 0; j < 22; j++) g[x][j] = sum[x] ^ (1 << j);
  out[x] = cnt;
}
void dfs2(int x) {
  if (!son[x]) {
    f[sum[x]] = dep[x];
    return;
  }
  for (int i = head[x]; i; i = ne[i]) {
    int y = to[i];
    if (y == son[x]) continue;
    dfs2(y);
    ans[x] = max(ans[x], ans[y]);
    for (int j = dfn[y]; j <= out[y]; j++) f[sum[rk[j]]] = 0;
  }
  dfs2(son[x]);
  ans[x] = max(ans[x], ans[son[x]]);
  for (int i = head[x]; i; i = ne[i]) {
    int y = to[i];
    if (y == son[x]) continue;
    for (int j = dfn[y]; j <= out[y]; j++) {
      for (int k = 0; k <= 22; k++) {
        if (f[g[rk[j]][k]])
          ans[x] = max(ans[x], f[g[rk[j]][k]] + dep[rk[j]] - dep[x] * 2);
      }
    }
    for (int j = dfn[y]; j <= out[y]; j++)
      f[sum[rk[j]]] = max(f[sum[rk[j]]], dep[rk[j]]);
  }
  for (int i = 0; i <= 22; i++) ans[x] = max(ans[x], f[g[x][i]] - dep[x]);
  f[sum[x]] = max(f[sum[x]], dep[x]);
}
int main() {
  scanf("%d", &n);
  for (int i = 2, x; i <= n; i++) {
    scanf("%d%s", &x, t + 1);
    add(x, i, t[1] - 'a');
  }
  dep[1] = 1;
  dfs1(1);
  dfs2(1);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
