Xenia the programmer has a tree consisting of n nodes. We will consider the tree nodes indexed from 1 to n. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.

The distance between two tree nodes v and u is the number of edges in the shortest path between v and u.

Xenia needs to learn how to quickly execute queries of two types:

  1. paint a specified blue node in red; 
  2. calculate which red node is the closest to the given one and print the shortest distance to the closest red node. 



Your task is to write a program which will execute the described queries.

Input

The first line contains two integers n and m (2 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of nodes in the tree and the number of queries. Next n - 1 lines contain the tree edges, the i-th line contains a pair of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — an edge of the tree.

Next m lines contain queries. Each query is specified as a pair of integers ti, vi (1 ≤ ti ≤ 2, 1 ≤ vi ≤ n). If ti = 1, then as a reply to the query we need to paint a blue node vi in red. If ti = 2, then we should reply to the query by printing the shortest distance from some red node to node vi.

It is guaranteed that the given graph is a tree and that all queries are correct.

Output

For each second type query print the reply in a single line.

Examples

Input

5 4
1 2
2 3
2 4
4 5
2 1
2 5
1 2
2 5


Output

0
3
2
#include <bits/stdc++.h>
using namespace std;
int d[100005];
vector<int> tree[100005];
int t, k = 0;
int arr[100005];
void solve() {
  for (int i = 0; i < k; i++) {
    int u = arr[i], siz = tree[u].size();
    for (int j = 0; j < siz; j++) {
      int v = tree[u][j];
      if (d[v] <= 1 + d[u]) continue;
      d[v] = 1 + d[u];
      arr[k++] = v;
    }
  }
  k = 0;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int u, v;
  fill(d, d + n + 1, INT_MAX);
  n--;
  while (n--) {
    scanf("%d%d", &u, &v);
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  d[1] = 0;
  arr[k++] = 1;
  while (m--) {
    scanf("%d%d", &t, &v);
    if (t == 1) {
      arr[k++] = v;
      d[v] = 0;
    } else {
      solve();
      printf("%d\n", d[v]);
    }
  }
  return 0;
}
