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 MAXN = 112345;
const int LOG = 18;
int dest[MAXN];
int q[MAXN * 32];
int len;
vector<vector<int> > tree;
int par[LOG][MAXN];
int mytime;
int tin[MAXN], tout[MAXN];
int rootdest[MAXN];
void dfs(int node, int pa) {
  par[0][node] = pa;
  tin[node] = mytime++;
  rootdest[node] = rootdest[pa] + 1;
  for (int i = 1; i < LOG; i++) {
    par[i][node] = par[i - 1][par[i - 1][node]];
  }
  for (int i = 0; i < tree[node].size(); i++) {
    int cur = tree[node][i];
    if (pa != cur) {
      dfs(cur, node);
    }
  }
  tout[node] = mytime++;
}
bool ispar(int u, int v) { return tin[u] <= tin[v] && tout[v] <= tout[u]; }
int lca(int u, int v) {
  if (ispar(u, v)) {
    return u;
  }
  if (ispar(v, u)) {
    return v;
  }
  for (int step = LOG - 1; step >= 0; step--) {
    if (!ispar(par[step][u], v)) {
      u = par[step][u];
    }
  }
  return par[0][u];
}
void bfs() {
  for (int i = 0; i < len; i++) {
    int cur = q[i];
    for (int i = 0; i < tree[cur].size(); i++) {
      int nxt = tree[cur][i];
      if (dest[cur] + 1 < dest[nxt]) {
        dest[nxt] = dest[cur] + 1;
        q[len++] = nxt;
      }
    }
  }
  len = 0;
}
int main() {
  int n, nq;
  scanf("%d%d", &n, &nq);
  tree.resize(n + 1);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  memset(dest, 63, sizeof dest);
  dfs(1, 1);
  dest[1] = 0;
  q[len++] = 1;
  bfs();
  while (nq--) {
    int t, v;
    scanf("%d%d", &t, &v);
    if (t == 1) {
      q[len++] = v;
      dest[v] = 0;
      if (len == 320) {
        bfs();
      }
    } else {
      int ans = dest[v];
      for (int i = 0; i < len; i++) {
        int cur = q[i];
        ans = min(ans, rootdest[cur] + rootdest[v] - 2 * rootdest[lca(cur, v)]);
      }
      printf("%d\n", ans);
    }
  }
}
