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;
vector<vector<int> > AdjList;
vector<int> dist;
vector<int> unused;
void update() {
  queue<int> now;
  for (int i = 0; i < unused.size(); i++) {
    dist[unused[i]] = 0;
    now.push(unused[i]);
  }
  while (!now.empty()) {
    int u = now.front();
    now.pop();
    for (int i = 0; i < AdjList[u].size(); i++) {
      int v = AdjList[u][i];
      if (dist[u] + 1 < dist[v]) {
        dist[v] = dist[u] + 1;
        now.push(v);
      }
    }
  }
  unused.clear();
}
int main() {
  int V, q;
  cin >> V >> q;
  AdjList.assign(V + 1, vector<int>());
  dist.assign(V + 1, -1);
  for (int x, y, i = 0; i < V - 1; i++) {
    cin >> x >> y;
    AdjList[x].push_back(y);
    AdjList[y].push_back(x);
  }
  dist[1] = 0;
  queue<int> now;
  now.push(1);
  while (!now.empty()) {
    int u = now.front();
    now.pop();
    for (int i = 0; i < AdjList[u].size(); i++) {
      int v = AdjList[u][i];
      if (dist[v] == -1) {
        dist[v] = dist[u] + 1;
        now.push(v);
      }
    }
  }
  for (int t, from, i = 0; i < q; i++) {
    cin >> t >> from;
    if (t == 1) {
      unused.push_back(from);
    } else {
      update();
      cout << dist[from] << endl;
    }
  }
  return 0;
}
