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 N = 112345;
int n, m;
vector<int> tree[N], c_tree[N];
int tree_sz[N], ans[N], parent[N];
bool ok[N];
map<int, map<int, int>> dist;
void calc_dist(int u, int c, int p = 0, int d = 0) {
  dist[c][u] = d;
  for (int v : tree[u]) {
    if (v != p and !ok[v]) {
      calc_dist(v, c, u, d + 1);
    }
  }
}
int get_sizes(int u, int p) {
  int sz = 0;
  for (int v : tree[u]) {
    if (v != p && !ok[v]) sz += get_sizes(v, u);
  }
  return tree_sz[u] = sz + 1;
}
int find_centroid(int u, int p, int sz) {
  for (int v : tree[u]) {
    if (v != p && !ok[v] && tree_sz[v] > sz / 2) return find_centroid(v, u, sz);
  }
  return u;
}
int build_ctree(int root) {
  int sz = get_sizes(root, 0);
  int centroid = find_centroid(root, 0, sz);
  ok[centroid] = true;
  calc_dist(centroid, centroid);
  for (int v : tree[centroid]) {
    if (!ok[v]) {
      int child_centroid = build_ctree(v);
      c_tree[centroid].emplace_back(child_centroid);
      parent[child_centroid] = centroid;
    }
  }
  return centroid;
}
void update(int x) {
  for (int u = x; u != 0; u = parent[u]) {
    ans[u] = min(ans[u], dist[u][x]);
  }
}
int query(int x) {
  int ret = 0x3f3f3f3f;
  for (int u = x; u != 0; u = parent[u]) {
    ret = min(ret, dist[u][x] + ans[u]);
  }
  return ret;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  memset(ans, 0x3f, sizeof ans);
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    tree[a].emplace_back(b);
    tree[b].emplace_back(a);
  }
  parent[build_ctree(1)] = 0;
  update(1);
  for (int i = 0; i < m; i++) {
    int t, x;
    cin >> t >> x;
    if (t == 1)
      update(x);
    else
      cout << query(x) << endl;
  }
  return 0;
}
