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 INF = 1e9;
vector<int> G[100005];
int vis[300005];
int anc[300005], pa[300005];
vector<int> dis[100005], nodes;
int sz[300005], maxsz[300005], d[300005];
int mn[300005], submn[300005], imp[300005];
void get_sz(int u) {
  nodes.push_back(u);
  vis[u] = 1;
  sz[u] = 1;
  maxsz[u] = 0;
  for (int v : G[u]) {
    if (!vis[v]) {
      get_sz(v);
      sz[u] += sz[v];
      maxsz[u] = max(sz[v], maxsz[u]);
    }
  }
}
void get_dis(int u, int p) {
  dis[u].push_back(d[u]);
  for (int v : G[u]) {
    if (!vis[v] && v != p) {
      d[v] = d[u] + 1;
      get_dis(v, u);
    }
  }
}
void get_cen(int u, int p) {
  nodes.clear();
  get_sz(u);
  int tot = nodes.size(), centroid = 0;
  for (int v : nodes) {
    vis[v] = 0;
    if (max(maxsz[v], tot - sz[v]) * 2 <= tot) centroid = v;
  }
  d[centroid] = 0;
  get_dis(centroid, centroid);
  pa[centroid] = p;
  vis[centroid] = 1;
  for (int v : G[centroid]) {
    if (!vis[v]) {
      get_cen(v, centroid);
    }
  }
}
void update(int u) {
  submn[u] = mn[u], mn[u] = 0, imp[u] = u;
  int x = pa[u], k = dis[u].size() - 2, last = u;
  while (x) {
    if (dis[u][k] <= mn[x]) {
      imp[x] = last, submn[x] = mn[x], mn[x] = dis[u][k];
    } else if (dis[u][k] < submn[x])
      submn[x] = dis[u][k];
    last = x, x = pa[x], k--;
  }
}
int query(int u) {
  int x = pa[u], k = dis[u].size() - 2, ans = mn[u], last = u;
  while (x) {
    if (imp[x] == last)
      ans = min(ans, submn[x] + dis[u][k]);
    else
      ans = min(ans, mn[x] + dis[u][k]);
    last = x, x = pa[x], k--;
  }
  return ans;
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  get_cen(1, 0);
  for (int i = 1; i <= n; i++) mn[i] = submn[i] = INF;
  update(1);
  while (q--) {
    int t, v;
    scanf("%d%d", &t, &v);
    if (t == 1)
      update(v);
    else
      printf("%d\n", query(v));
  }
}
