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 = 1e5 + 20;
int n, m;
map<int, int> Mp[N];
int mn[N];
int f[N], sz[N], Rt, vis[N], all;
vector<int> P[N], E[N];
void Get_Rt(int x, int F) {
  sz[x] = 1, f[x] = 0;
  for (int i = 0; i < E[x].size(); i++) {
    int V = E[x][i];
    if (V == F || vis[V]) continue;
    Get_Rt(V, x);
    sz[x] += sz[V];
    f[x] = max(f[x], sz[V]);
  }
  f[x] = max(f[x], all - sz[x]);
  if (f[x] < f[Rt]) Rt = x;
}
void Cal(int x, int f, int Rt, int dis) {
  Mp[x][Rt] = dis;
  mn[Rt] = 0x3f3f3f3f;
  P[x].push_back(Rt);
  for (int i = 0; i < E[x].size(); i++) {
    int V = E[x][i];
    if (V == f || vis[V]) continue;
    Cal(V, x, Rt, dis + 1);
  }
}
void dfs(int x) {
  vis[x] = 1;
  Cal(x, -1, x, 0);
  for (int i = 0; i < E[x].size(); i++) {
    int V = E[x][i];
    if (vis[V]) continue;
    all = sz[V];
    Rt = 0;
    Get_Rt(V, x);
    dfs(Rt);
  }
}
void Update(int x) {
  for (int i = 0; i < P[x].size(); i++) {
    int Fa = P[x][i];
    mn[Fa] = min(mn[Fa], Mp[x][Fa]);
  }
}
void Query(int x) {
  int ans = 0x3f3f3f3f;
  for (int i = 0; i < P[x].size(); i++) {
    int Fa = P[x][i];
    ans = min(ans, mn[Fa] + Mp[x][Fa]);
  }
  printf("%d\n", ans);
}
void PreSolve() {
  all = n;
  Rt = 0;
  f[0] = 0x3f3f3f3f;
  Get_Rt(1, -1);
  dfs(Rt);
  Update(1);
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 2; i <= n; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    E[a].push_back(b);
    E[b].push_back(a);
  }
  PreSolve();
  for (int i = 1; i <= m; i++) {
    int t, x;
    scanf("%d %d", &t, &x);
    if (t == 1)
      Update(x);
    else
      Query(x);
  }
  return 0;
}
