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 INF = (int)(1e8);
vector<int> g[100100];
int depths[100100];
int up[100100][18];
int tin[100100], tout[100100];
int timer = 1;
void dfs(int v, int p, int h) {
  depths[v] = h;
  tin[v] = timer++;
  up[v][0] = p;
  for (int i = 1; i < 18; i++) up[v][i] = up[up[v][i - 1]][i - 1];
  for (size_t i = 0; i < g[v].size(); i++)
    if (g[v][i] != p) dfs(g[v][i], v, h + 1);
  tout[v] = timer++;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = 17; i >= 0; i--)
    if (!upper(up[a][i], b)) a = up[a][i];
  return up[a][0];
}
int distance(int a, int b) {
  return depths[a] + depths[b] - 2 * depths[lca(a, b)];
}
int n, m;
bool color[100100];
bool used[100100];
int dist[100100];
int block, block_cnt;
void bfs() {
  queue<int> q;
  for (int i = 0; i < n; i++) {
    if (color[i]) {
      dist[i] = 0;
      used[i] = true;
      q.push(i);
    } else {
      dist[i] = INF;
      used[i] = false;
    }
  }
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    for (int i = 0; i < g[v].size(); i++) {
      int to = g[v][i];
      if (!used[to]) {
        dist[to] = 1 + dist[v];
        used[to] = true;
        q.push(to);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  scanf("%d %d\n", &n, &m);
  int a, b;
  for (int i = 1; i < n; i++) {
    scanf("%d %d\n", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0, 0);
  color[0] = true;
  block = 1 + (int)sqrt(m + 0.0);
  block_cnt = (m / block) + (m % block);
  for (int k = 0; k < block_cnt; k++) {
    int l = block * k;
    int r = min(l + block, m);
    vector<int> red;
    bfs();
    for (int i = l; i < r; i++) {
      int t, v;
      scanf("%d %d\n", &t, &v);
      --v;
      if (t == 1) {
        red.push_back(v);
        color[v] = true;
        dist[v] = 0;
      } else {
        for (int j = 0; j < red.size(); j++)
          dist[v] = min(dist[v], distance(red[j], v));
        cout << dist[v] << endl;
      }
    }
  }
}
