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 maxn = 1e5 + 6, mod = 1e9 + 7;
int n, m, par[20][maxn], h[maxn], pw2[30], ans[maxn], d[maxn], st[maxn],
    ft[maxn], tme;
bool mrk[maxn];
vector<int> adj[maxn], vec;
void DFS(int u) {
  mrk[u] = 1;
  st[u] = tme++;
  for (int i = 1; i < 17; i++) {
    int v = par[i - 1][u];
    if (v == 0 || par[i - 1][v] == 0) break;
    par[i][u] = par[i - 1][v];
  }
  for (int j : adj[u]) {
    if (mrk[j]) continue;
    par[0][j] = u;
    h[j] = h[u] + 1;
    DFS(j);
  }
  ft[u] = tme;
}
int lca(int u, int v) {
  if (h[u] > h[v]) swap(u, v);
  int dis = h[v] - h[u];
  for (int i = 16; i >= 0; i--) {
    if (dis & pw2[i]) v = par[i][v];
  }
  if (u == v) return u;
  for (int i = 16; i >= 0; i--) {
    if (par[i][v] != par[i][u]) {
      u = par[i][u];
      v = par[i][v];
    }
  }
  return par[0][u];
}
void bfs() {
  queue<int> q;
  for (int j : vec) {
    q.push(j);
    d[j] = 0;
    mrk[j] = 1;
  }
  while (q.size()) {
    int v = q.front();
    q.pop();
    ans[v] = min(ans[v], d[v]);
    for (int j : adj[v]) {
      if (mrk[j]) continue;
      q.push(j);
      d[j] = d[v] + 1;
      mrk[j] = 1;
    }
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  pw2[0] = 1;
  for (int i = 1; i < 20; i++) pw2[i] = pw2[i - 1] * 2ll;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  DFS(1);
  memset(mrk, 0, sizeof mrk);
  fill(ans, ans + n + 1, n);
  int T = 90;
  vec.push_back(1);
  while (m--) {
    int t, u;
    scanf("%d%d", &t, &u);
    if (t == 2) {
      if (ans[u] > 0)
        for (int i : vec) {
          if (((st[u] < st[i]) && ft[u] >= ft[i]) ||
              ((st[u] > st[i]) && ft[u] <= ft[i]))
            ans[u] = min(ans[u], abs(h[i] - h[u]));
          else
            ans[u] = min(ans[u], h[u] + h[i] - 2 * h[lca(u, i)]);
        }
      cout << ans[u] << "\n";
      continue;
    } else {
      vec.push_back(u);
      if (vec.size() == T) {
        memset(mrk, 0, sizeof mrk);
        bfs();
        vec.clear();
      }
    }
  }
}
