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 = 200010;
const int M = 25;
const int mod = 1e9 + 7;
int n, m;
vector<int> tree[N];
int level[N], in[N];
vector<int> euler;
bitset<N> v;
int spt[N + N][M];
void dfs(int id) {
  in[id] = (int)euler.size(), v[id] = 1;
  euler.push_back(id);
  for (auto &i : tree[id]) {
    if (!v[i]) level[i] = level[id] + 1, dfs(i);
    euler.push_back(id);
  }
  euler.push_back(id);
}
void preLCA() {
  for (int i = 0; i < euler.size(); i++) spt[i][0] = euler[i];
  for (int j = 1; (1 << j) <= euler.size(); j++)
    for (int i = 0; i + (1 << j) - 1 < euler.size(); i++) {
      spt[i][j] = (level[spt[i][j - 1]] < level[spt[i + (1 << (j - 1))][j - 1]])
                      ? spt[i][j - 1]
                      : spt[i + (1 << (j - 1))][j - 1];
    }
}
int LCA(int i, int j) {
  i = in[i], j = in[j];
  if (i > j) swap(i, j);
  int k = log2(j - i + 1.0);
  return (level[spt[i][k]] < level[spt[j - (1 << k) + 1][k]])
             ? spt[i][k]
             : spt[j - (1 << k) + 1][k];
}
int root;
int sz[N];
int centroidp[N];
int big;
int dfsC(int id, int p = -1) {
  sz[id] = 1;
  for (auto &i : tree[id])
    if (!v[i] && i != p) sz[id] += dfsC(i, id);
  return sz[id];
}
int centroid(int id, int p = -1) {
  for (auto &i : tree[id])
    if (!v[i] && sz[i] + sz[i] > big && i != p) return centroid(i, id);
  return id;
}
int centroidDecomp(int id) {
  dfsC(id);
  big = sz[id];
  int ans = centroid(id);
  v[ans] = 1;
  for (auto &i : tree[ans])
    if (!v[i]) {
      int who = centroidDecomp(i);
      centroidp[who] = ans;
    }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int x, y;
  int dist[N];
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    tree[x].push_back(y);
    tree[y].push_back(x);
  }
  dfs(1);
  preLCA();
  v.reset();
  big = n;
  root = centroidDecomp(1);
  for (int i = 0; i <= n; i++) dist[i] = mod;
  {
    y = 1;
    int who = y;
    dist[y] = 0;
    while (who != root) {
      who = centroidp[who];
      dist[who] =
          min(dist[who], level[y] + level[who] - (level[LCA(y, who)] << 1));
    }
  }
  while (m--) {
    cin >> x >> y;
    if (x == 1) {
      int who = y;
      dist[y] = 0;
      while (who != root) {
        who = centroidp[who];
        dist[who] =
            min(dist[who], level[y] + level[who] - (level[LCA(y, who)] << 1));
      }
    } else {
      int ans = dist[y];
      int who = y;
      while (who != root) {
        who = centroidp[who];
        ans = min(
            ans, level[y] + level[who] - (level[LCA(y, who)] << 1) + dist[who]);
      }
      cout << ans << '\n';
    }
  }
}
