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<int> G[100000];
int depth[100000], par[17][100000], d[100000];
void dfs(int v, int p, int dep) {
  par[0][v] = p;
  depth[v] = d[v] = dep;
  for (int i = 0; i < G[v].size(); i++) {
    if (G[v][i] != p) {
      dfs(G[v][i], v, dep + 1);
    }
  }
}
void init(int n) {
  memset(par, -1, sizeof(par));
  dfs(0, -1, 0);
  for (int i = 0; i + 1 < 17; i++) {
    for (int j = 0; j < n; j++) {
      if (~par[i][j]) par[i + 1][j] = par[i][par[i][j]];
    }
  }
}
int lca(int u, int v) {
  if (depth[u] < depth[v]) swap(u, v);
  for (int i = 0; i < 17; i++) {
    if ((depth[u] - depth[v]) >> i & 1) {
      u = par[i][u];
    }
  }
  if (u == v) return (u);
  for (int i = 16; i >= 0; i--) {
    if (par[i][u] != par[i][v]) {
      u = par[i][u];
      v = par[i][v];
    }
  }
  return (par[0][u]);
}
void bfs(vector<int> v) {
  bool vis[100000];
  fill(vis, vis + 100000, 0);
  queue<pair<int, int> > q;
  for (int i = 0; i < v.size(); i++) {
    q.push(make_pair(v[i], 0));
    vis[v[i]] = true;
  }
  for (; q.size(); q.pop()) {
    pair<int, int> x = q.front();
    if (x.second < d[x.first]) d[x.first] = x.second;
    for (int i = 0; i < G[x.first].size(); i++) {
      if (!vis[G[x.first][i]]) {
        vis[G[x.first][i]] = true;
        q.push(make_pair(G[x.first][i], x.second + 1));
      }
    }
  }
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    G[--a].push_back(--b);
    G[b].push_back(a);
  }
  init(n);
  vector<int> block;
  block.push_back(0);
  for (int i = 0; i < m; i++) {
    if (i % 317 == 0) {
      bfs(block);
      block.clear();
    }
    int q, v;
    scanf("%d %d", &q, &v);
    if (q == 1) {
      block.push_back(--v);
    } else {
      int ret = d[--v];
      for (int j = 0; j < block.size(); j++) {
        ret =
            min(ret, depth[v] + depth[block[j]] - 2 * depth[lca(v, block[j])]);
      }
      printf("%d\n", ret);
    }
  }
  return (0);
}
