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>
const int N = 100010;
const long long INF = 1ll << 60;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
int n, m;
std::vector<int> g[N];
int st[N], cnt;
int dep[N];
int id[N << 2 | 1][21];
int lg2[N << 2 | 1];
void Dfs(int u, int f) {
  st[u] = ++cnt;
  dep[u] = dep[f] + 1;
  id[cnt][0] = u;
  for (size_t i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (v == f) continue;
    Dfs(v, u);
    id[++cnt][0] = u;
  }
}
void Rmq() {
  for (int i = 2; i <= cnt; ++i) lg2[i] = lg2[i >> 1] + 1;
  for (int j = 1; j <= 20; ++j) {
    for (int i = 1; (i + (1 << j) - 1) <= cnt; ++i) {
      int r = i + (1 << (j - 1));
      id[i][j] =
          dep[id[i][j - 1]] < dep[id[r][j - 1]] ? id[i][j - 1] : id[r][j - 1];
    }
  }
}
int Lca(int u, int v) {
  int l = st[u];
  int r = st[v];
  if (l > r) std::swap(l, r);
  int k = lg2[r - l + 1];
  int m = r - (1 << k) + 1;
  return dep[id[l][k]] < dep[id[m][k]] ? id[l][k] : id[m][k];
}
std::vector<int> block;
int sn;
bool red[N];
std::queue<int> q;
int d[N];
void Bfs() {
  for (int i = 1; i <= n; ++i) {
    if (red[i]) {
      d[i] = 0;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (size_t i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      if (d[v] > d[u] + 1) {
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  sn = (int)sqrt(n);
  memset(d, inf, sizeof(d));
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  Dfs(1, 0);
  Rmq();
  red[1] = true;
  Bfs();
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    if (x == 1) {
      block.push_back(y);
      red[y] = true;
    } else {
      if (block.size() > sn) {
        block.clear();
        Bfs();
        printf("%d\n", d[y]);
      } else {
        int ans = d[y];
        for (size_t i = 0; i < block.size(); ++i) {
          ans =
              std::min(ans, dep[y] + dep[block[i]] - 2 * dep[Lca(y, block[i])]);
        }
        printf("%d\n", ans);
      }
    }
  }
  return 0;
}
