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 long long INF = 1e18;
const int maxn = 2e5 + 10;
const long long MOD = 1000000007;
const double EPS = 1e-10;
const double Pi = acos(-1.0);
vector<int> block;
int dp[maxn];
const int MAX_V = 1e5 + 10;
const int MAX_LOG_V = 18;
vector<int> G[MAX_V];
int root;
int parent[MAX_LOG_V][MAX_V];
int depth[MAX_V];
void dfs(int v, int p, int d) {
  parent[0][v] = p;
  depth[v] = d;
  for (int i = 0; i < G[v].size(); i++)
    if (G[v][i] != p) dfs(G[v][i], v, d + 1);
}
void init(int V) {
  dfs(root, -1, 0);
  for (int k = 0; k + 1 < MAX_LOG_V; k++)
    for (int v = 1; v <= V; v++)
      if (parent[k][v] < 0)
        parent[k + 1][v] = -1;
      else
        parent[k + 1][v] = parent[k][parent[k][v]];
}
int lca(int u, int v) {
  if (depth[u] > depth[v]) swap(u, v);
  for (int k = 0; k < MAX_LOG_V; k++)
    if ((depth[v] - depth[u]) >> k & 1) v = parent[k][v];
  if (u == v) return u;
  for (int k = MAX_LOG_V - 1; k >= 0; k--)
    if (parent[k][u] != parent[k][v]) u = parent[k][u], v = parent[k][v];
  return parent[0][u];
}
void bfs() {
  queue<int> Q;
  for (int i = 0; i < block.size(); i++) Q.push(block[i]), dp[block[i]] = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = 0; i < G[u].size(); i++) {
      int v = G[u][i];
      if (dp[v] > dp[u] + 1) dp[v] = dp[u] + 1, Q.push(v);
    }
  }
  block.clear();
}
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 0; i <= n; i++) G[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    root = 1;
    init(n);
    for (int i = 1; i <= n; i++) dp[i] = 0x3f3f3f3f;
    block.clear();
    block.push_back(1);
    int mq = sqrt(m);
    for (int i = 0; i < m; i++) {
      int op, u;
      scanf("%d%d", &op, &u);
      if (op == 1) {
        block.push_back(u);
        if (block.size() == mq) bfs();
      } else {
        int ans = dp[u];
        for (int i = 0; i < block.size(); i++) {
          int v = block[i];
          ans = min(ans, depth[u] + depth[v] - 2 * depth[lca(u, v)]);
        }
        printf("%d\n", ans);
      }
    }
  }
  return 0;
}
