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 n, m;
vector<int> g[100001];
bool is_centroid[100001];
int centroid_parent[100001];
int mn[100001];
int sz[100001];
int level[100001];
int centroid_root;
map<pair<int, int>, int> dist;
void dfs(int nd, int p, int c) {
  sz[nd] = 1;
  if (p == c) {
    level[nd] = 0;
  } else {
    level[nd] = level[p] + 1;
  }
  dist[make_pair(nd, c)] = level[nd] + 1;
  for (int i = 0; i < g[nd].size(); i++) {
    int z = g[nd][i];
    if (z != p && !is_centroid[z]) {
      dfs(z, nd, c);
      sz[nd] += sz[z];
    }
  }
}
int pd;
int find_centroid(int u, int ts) {
  for (int i = 0; i < g[u].size(); i++) {
    int z = g[u][i];
    if (is_centroid[z] || sz[z] > sz[u]) continue;
    if (sz[z] > ts / 2) return find_centroid(z, ts);
  }
  return u;
}
int get_centroid(int nd, int p) {
  dfs(nd, p, p);
  int u = find_centroid(nd, sz[nd]);
  is_centroid[u] = true;
  centroid_parent[u] = p;
  for (int i = 0; i < g[u].size(); i++) {
    int z = g[u][i];
    if (!is_centroid[z]) {
      get_centroid(z, u);
    }
  }
  return u;
}
void centroid_decompose() {
  fill(is_centroid, is_centroid + 100001, false);
  centroid_root = get_centroid(1, -1);
}
void debug() {
  for (int i = 1; i <= n; i++) {
    cout << i << " " << centroid_parent[i] << " " << mn[i] << endl;
  }
}
void redify(int u) {
  int x = u;
  while (u != -1) {
    mn[u] = min(dist[make_pair(x, u)], mn[u]);
    u = centroid_parent[u];
  }
}
int solve(int u) {
  int res = INT_MAX / 3;
  int x = u;
  while (u != -1) {
    res = min(res, dist[make_pair(x, u)] + mn[u]);
    u = centroid_parent[u];
  }
  return res;
}
int main() {
  int u, v;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  fill(mn, mn + 100001, INT_MAX / 3);
  centroid_decompose();
  redify(1);
  while (m--) {
    scanf("%d %d", &u, &v);
    if (u == 1) {
      redify(v);
    } else {
      printf("%d\n", solve(v));
    }
  }
  return 0;
}
