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 = 100010, DE = 20, D = 250, inf = 1e9;
int n, m, d[N], b[N], t[N], x[N], q[N][DE], sol[N];
vector<int> v[N];
queue<int> qu;
void dfs(int c, int p) {
  d[c] = d[p] + 1;
  q[c][0] = p;
  for (int i = 1; i < DE; i++) q[c][i] = q[q[c][i - 1]][i - 1];
  for (auto i : v[c])
    if (i != p) dfs(i, c);
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = DE - 1; i >= 0; i--)
    if (d[x] - d[y] >> i & 1) x = q[x][i];
  if (x == y) return x;
  for (int i = DE - 1; i >= 0; i--)
    if (q[x][i] != q[y][i]) x = q[x][i], y = q[y][i];
  return q[x][0];
}
int dist(int x, int y) { return d[x] + d[y] - 2 * d[lca(x, y)]; }
void bfs() {
  for (int i = 1; i <= n; i++) sol[i] = inf;
  for (int i = 1; i <= n; i++)
    if (b[i]) sol[i] = 0, qu.push(i);
  while (!qu.empty()) {
    int c = qu.front();
    qu.pop();
    for (auto i : v[c])
      if (sol[i] == inf) sol[i] = sol[c] + 1, qu.push(i);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  b[1] = true;
  for (int i = 0; i < m; i++) {
    if (!(i % D)) bfs();
    scanf("%d %d", &t[i], &x[i]);
    if (t[i] == 1)
      b[x[i]] = true;
    else {
      int z = sol[x[i]];
      for (int j = i - 1; j >= 0 && i / D == j / D; j--)
        if (t[j] == 1) z = min(z, dist(x[i], x[j]));
      printf("%d\n", z);
    }
  }
  return 0;
}
