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 const N = 200200;
int const LOG = 19;
vector<int> g[N];
int cdep[N], cpar[N];
int under[N];
int tin[N], depth[N];
int sp[LOG][N], logs[N], timer = 0;
void dfs(int v, int par, int dep) {
  depth[v] = dep;
  sp[0][tin[v] = timer++] = dep;
  for (int to : g[v]) {
    if (to != par) {
      dfs(to, v, dep + 1);
      sp[0][timer++] = dep;
    }
  }
}
int dist(int a, int b) {
  int x = min(tin[a], tin[b]), y = max(tin[a], tin[b]) + 1;
  int i = logs[y - x];
  return depth[a] + depth[b] - 2 * min(sp[i][x], sp[i][y - (1 << i)]);
}
void calc_sizes(int v, int par) {
  under[v] = 1;
  for (int to : g[v]) {
    if (to != par && cdep[to] < 0) {
      calc_sizes(to, v);
      under[v] += under[to];
    }
  }
}
void build_cd(int v, int dep, int par) {
  calc_sizes(v, -1);
  for (int did = 1, size = under[v]; did;) {
    did = 0;
    for (int to : g[v]) {
      if (cdep[to] < 0 && under[to] < under[v] && under[to] * 2 > size) {
        v = to;
        did = 1;
      }
    }
  }
  cdep[v] = dep;
  cpar[v] = par;
  for (int to : g[v])
    if (cdep[to] < 0) build_cd(to, dep + 1, v);
}
vector<pair<int, int>> dp[N];
void add(int v) {
  for (int pc = -1, c = v; c >= 0; pc = c, c = cpar[c]) {
    int pos = dp[c].size();
    dp[c].emplace_back(dist(v, c), pc);
    while (pos && dp[c][pos] < dp[c][pos - 1]) {
      swap(dp[c][pos], dp[c][pos - 1]);
      --pos;
    }
    if ((int)dp[c].size() > 2) dp[c].pop_back();
  }
}
int ans(int v) {
  int ret = N;
  for (int pc = -2, c = v; c >= 0; pc = c, c = cpar[c]) {
    int curd = dist(v, c);
    for (auto x : dp[c])
      if (x.second != pc) ret = min(ret, curd + x.first);
  }
  return ret;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, 0, 0);
  for (int i = 1; (1 << i) <= timer; ++i) {
    int len = 1 << i;
    for (int j = 0; j + len <= timer; ++j) {
      sp[i][j] = min(sp[i - 1][j], sp[i - 1][j + len / 2]);
    }
  }
  logs[1] = 0;
  for (int i = 2; i <= timer; ++i) logs[i] = 1 + logs[i / 2];
  fill(cdep, cdep + n, -1);
  build_cd(0, 0, -1);
  add(0);
  for (int i = 0; i < m; ++i) {
    int t, v;
    scanf("%d%d", &t, &v);
    --v;
    if (t == 1) {
      add(v);
    } else {
      cout << ans(v) << '\n';
    }
  }
}
