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> graph[100005];
int dp[20][100005], level[100005], sum[100005], parent[100005], done[100005],
    answer[100005];
void dfs0(int v) {
  for (int i = 0; i < graph[v].size(); i++) {
    int u = graph[v][i];
    if (u != dp[0][v]) {
      dp[0][u] = v;
      level[u] = level[v] + 1;
      dfs0(u);
    }
  }
}
void preprocess(int n) {
  dp[0][1] = 1;
  dfs0(1);
  for (int i = 1; i < 20; i++)
    for (int j = 1; j <= n; j++) dp[i][j] = dp[i - 1][dp[i - 1][j]];
}
int lca(int a, int b) {
  if (level[a] > level[b]) swap(a, b);
  int d = level[b] - level[a];
  for (int i = 0; i < 20; i++)
    if (d & (1 << i)) b = dp[i][b];
  if (a == b) return a;
  for (int i = 20 - 1; i >= 0; i--)
    if (dp[i][a] != dp[i][b]) a = dp[i][a], b = dp[i][b];
  return dp[0][a];
}
inline int dist(int a, int b) {
  return level[a] + level[b] - 2 * level[lca(a, b)];
}
int cnt;
void dfs1(int v, int p) {
  sum[v] = 1;
  cnt++;
  for (int i = 0; i < graph[v].size(); i++) {
    int u = graph[v][i];
    if (u != p && !done[u]) {
      dfs1(u, v);
      sum[v] += sum[u];
    }
  }
}
int dfs2(int v, int p) {
  for (int i = 0; i < graph[v].size(); i++) {
    int u = graph[v][i];
    if (u != p && sum[u] > cnt / 2 && !done[u]) return dfs2(u, v);
  }
  return v;
}
void decompose(int v, int p) {
  cnt = 0;
  dfs1(v, v);
  int centroid = dfs2(v, v);
  if (p == -1) p = centroid;
  parent[centroid] = p;
  done[centroid] = true;
  for (int i = 0; i < graph[centroid].size(); i++) {
    int u = graph[centroid][i];
    if (!done[u]) decompose(u, centroid);
  }
}
void update(int v) {
  int x = v;
  while (true) {
    answer[x] = min(answer[x], dist(x, v));
    if (x == parent[x]) break;
    x = parent[x];
  }
}
int query(int v) {
  int x = v, ans = 100005;
  while (true) {
    ans = min(ans, dist(x, v) + answer[x]);
    if (x == parent[x]) break;
    x = parent[x];
  }
  return ans;
}
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);
    graph[a].push_back(b), graph[b].push_back(a);
  }
  for (int i = 1; i < 100005; i++) answer[i] = 100005;
  preprocess(n);
  decompose(1, -1);
  update(1);
  for (int i = 0; i < m; i++) {
    int t, v;
    scanf("%d %d", &t, &v);
    if (t == 1)
      update(v);
    else
      printf("%d\n", query(v));
  }
  return 0;
}
