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>
long double PI = 3.141592653589793238462643383279502884197;
using namespace std;
vector<int> vec[100001];
int ch[100001], pa[100001], mn[100001], dist[20][100001], dep[100001];
bool use[100001];
int getSize(int cur, int p) {
  ch[cur] = 1;
  for (int v : vec[cur]) {
    if (p != v && !use[v]) ch[cur] += getSize(v, cur);
  }
  return ch[cur];
}
int centroid(int cur, int p, int sz) {
  for (int v : vec[cur]) {
    if (p != v && !use[v] && ch[v] > sz) return centroid(v, cur, sz);
  }
  return cur;
}
void dfs(int cur, int p, int dd, int de) {
  dist[de][cur] = dd;
  for (int v : vec[cur]) {
    if (p != v && !use[v]) dfs(v, cur, dd + 1, de);
  }
}
int dq(int cur, int de) {
  int nx = centroid(cur, cur, getSize(cur, cur) / 2);
  use[nx] = true;
  dep[nx] = de;
  for (int v : vec[nx]) {
    if (use[v]) continue;
    dfs(v, nx, 1, de);
    pa[dq(v, de + 1)] = nx;
  }
  return nx;
}
void update(int cur) {
  int c = cur;
  for (int de = dep[cur]; de >= 0; de--, cur = pa[cur]) {
    mn[cur] = min(mn[cur], dist[de][c]);
  }
}
int getAns(int cur) {
  int ret = 10000000, c = cur;
  for (int de = dep[cur]; de >= 0; de--, cur = pa[cur]) {
    ret = min(ret, mn[cur] + dist[de][c]);
  }
  return ret;
}
int main() {
  int n, m, a, b, cm, x;
  scanf("%d %d", &n, &m);
  fill(mn, mn + n + 1, 10000000);
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &a, &b);
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  dq(1, 0);
  update(1);
  while (m--) {
    scanf("%d %d", &cm, &x);
    if (cm == 1) {
      update(x);
    } else if (cm == 2) {
      printf("%d\n", getAns(x));
    }
  }
  return 0;
}
