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;
int n, m, num[100005], lev[100005], par[100005][23 + 3], cen_par[100005];
long long dist[100005];
bool chk[100005];
vector<int> net[100005], cen[100005];
void make_num(int c, int p) {
  int y = 1;
  for (int i : net[c]) {
    if (i == p) continue;
    lev[i] = lev[c] + 1;
    par[i][0] = c;
    make_num(i, c);
    y += num[i];
  }
  num[c] = y;
  return;
}
int find_cen(int c) {
  int y = 1, mx = -1, mxi;
  for (int i : net[c]) {
    if (chk[i]) continue;
    if (mx < num[i]) {
      mx = num[i];
      mxi = i;
    }
    y += num[i];
  }
  if (mx <= y / 2) return c;
  num[c] = y - mx;
  return find_cen(mxi);
}
int make_tree(int c) {
  int x = find_cen(c);
  chk[x] = true;
  for (int i : net[x]) {
    if (chk[i]) continue;
    int t = make_tree(i);
    cen[x].emplace_back(t);
    cen_par[t] = x;
  }
  return x;
}
long long get_dist(int x, int y) {
  long long cnt = 0;
  if (lev[x] > lev[y]) swap(x, y);
  for (int i = 23; i >= 0; i--) {
    if ((lev[y] - lev[x]) >= (1 << i)) {
      y = par[y][i];
      cnt += (1 << i);
    }
  }
  if (x == y) return cnt;
  for (int i = 23; i >= 0; i--) {
    if (par[x][i] != par[y][i]) {
      x = par[x][i];
      y = par[y][i];
      cnt += (2 << i);
    }
  }
  return cnt + 2;
}
void update(int x) {
  int here = x;
  dist[here] = 0;
  while (x) {
    dist[x] = min(dist[x], get_dist(x, here));
    x = cen_par[x];
  }
  return;
}
long long read(int x) {
  long long ret = 2147483646, here = x;
  while (x) {
    ret = min(ret, dist[x] + get_dist(x, here));
    x = cen_par[x];
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0, t1, t2, c; i < n - 1; i++) {
    cin >> t1 >> t2;
    net[t1].emplace_back(t2);
    net[t2].emplace_back(t1);
  }
  make_num(1, -1);
  for (int j = 1; j <= 23; j++) {
    for (int i = 1; i <= n; i++) {
      par[i][j] = par[par[i][j - 1]][j - 1];
    }
  }
  find_cen(1);
  make_tree(1);
  for (int i = 1; i <= n; i++) dist[i] = 2147483646;
  update(1);
  for (int i = 0, t1, t2; i < m; i++) {
    cin >> t1 >> t2;
    if (t1 == 1) {
      update(t2);
    }
    if (t1 == 2) {
      cout << read(t2) << "\n";
    }
  }
}
