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 + 7;
const int BASE = 1e9 + 7;
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
const int N = 1e5 + 1;
set<int> a[N];
int n, m, u, v, total;
int F[N], pa[21][N], H[N];
int parent[N], ans[N], type;
void DFS(int p, int u) {
  for (__typeof(a[u].begin()) it = a[u].begin(); it != a[u].end(); it++) {
    int v = *it;
    if (v == p) continue;
    pa[0][v] = u;
    H[v] = H[u] + 1;
    DFS(u, v);
  }
}
int LCA(int u, int v) {
  if (H[u] < H[v]) swap(u, v);
  int diff = log2(H[u]);
  for (int i = diff; i >= 0; i--) {
    if (H[u] - (1 << i) >= H[v]) {
      u = pa[i][u];
    }
  }
  if (u == v) return u;
  diff = log2(H[u]);
  for (int i = diff; i >= 0; i--) {
    if (pa[i][u] != pa[i][v]) {
      u = pa[i][u];
      v = pa[i][v];
    }
  }
  return pa[0][u];
}
int DIS(int u, int v) {
  int x = LCA(u, v);
  return H[u] + H[v] - 2 * H[x];
}
void dfs_subtree(int p, int u) {
  F[u] = 1;
  total++;
  for (__typeof(a[u].begin()) it = a[u].begin(); it != a[u].end(); it++) {
    int v = *it;
    if (v == p) continue;
    dfs_subtree(u, v);
    F[u] += F[v];
  }
}
int FindCentroid(int p, int u) {
  for (__typeof(a[u].begin()) it = a[u].begin(); it != a[u].end(); it++) {
    int v = *it;
    if (v == p) continue;
    if (F[v] > total / 2) {
      return FindCentroid(u, v);
    }
  }
  return u;
}
void Centroid_decompose(int p, int u) {
  total = 0;
  dfs_subtree(0, u);
  int centroid = FindCentroid(0, u);
  if (p == 0) p = centroid;
  parent[centroid] = p;
  for (__typeof(a[centroid].begin()) it = a[centroid].begin();
       it != a[centroid].end(); it++) {
    int v = *it;
    a[v].erase(centroid);
    Centroid_decompose(centroid, v);
  }
  a[centroid].clear();
}
void Update(int u) {
  int x = u;
  while (true) {
    ans[x] = min(ans[x], DIS(u, x));
    if (x == parent[x]) break;
    x = parent[x];
  }
}
int Query(int u) {
  int x = u;
  int res = INF;
  while (true) {
    res = min(res, DIS(u, x) + ans[x]);
    if (x == parent[x]) return res;
    x = parent[x];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    a[v].insert(u);
    a[u].insert(v);
  }
  DFS(0, 1);
  for (int i = 1; (1 << i) <= n; i++) {
    for (int j = 1; j <= n; j++) {
      pa[i][j] = pa[i - 1][pa[i - 1][j]];
    }
  }
  Centroid_decompose(0, 1);
  for (int i = 1; i <= n; i++) ans[i] = INF;
  Update(1);
  while (m--) {
    cin >> type >> u;
    if (type == 1)
      Update(u);
    else {
      cout << Query(u) << '\n';
    }
  }
  return 0;
}
