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;
const int maxn = 100100;
vector<int> t[maxn];
bool c[maxn];
queue<int> q;
int dis[maxn];
int n, m;
void BFS() {
  int u, v;
  while (!q.empty()) {
    u = q.front();
    q.pop();
    for (vector<int>::iterator i = t[u].begin(); i != t[u].end(); i++) {
      v = *i;
      if (dis[v] > dis[u] + 1) {
        dis[v] = dis[u] + 1;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d %d\n", &n, &m);
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d %d\n", &u, &v);
    t[u].push_back(v);
    t[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) dis[i] = n << 1;
  c[1] = 1;
  q.push(1);
  dis[1] = 0;
  for (int o = 1; o <= m; o++) {
    scanf("%d %d", &u, &v);
    if (u == 1) {
      q.push(v);
      dis[v] = 0;
    } else {
      BFS();
      printf("%d\n", dis[v]);
    }
  }
  return 0;
}
