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 = 1e5 + 10, INF = 1e9;
int par[MAXN][30], h[MAXN];
vector<int> ne[MAXN];
int n, q;
void dfs(int v) {
  for (int i = 1; i < 30; ++i) par[v][i] = par[par[v][i - 1]][i - 1];
  for (auto i : ne[v])
    if (i != par[v][0]) {
      h[i] = h[v] + 1;
      par[i][0] = v;
      dfs(i);
    }
}
int dis(int v, int u) {
  int b = 29, ans = abs(h[u] - h[v]);
  if (h[v] < h[u]) swap(u, v);
  while (h[u] != h[v]) {
    while (b && h[v] - (1 << b) < h[u]) {
      --b;
    }
    v = par[v][b];
  }
  b = 29;
  int u1 = u, v1 = v;
  while (u1 != v1) {
    while (b && par[u1][b] == par[v1][b]) --b;
    u1 = par[u1][b], v1 = par[v1][b];
  }
  return ans + abs(h[u1] - h[u]) + abs(h[v1] - h[v]);
}
vector<int> D;
vector<int> red;
bitset<MAXN> b;
void upd() {
  queue<int> q;
  for (int i = 0; i <= n; ++i) b[i] = false;
  for (auto i : red) b[i] = true;
  for (auto i : red) q.push(i);
  for (auto i : red) D[i] = 0;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (auto i : ne[now])
      if (!b[i]) {
        b[i] = true;
        q.push(i);
        D[i] = D[now] + 1;
      }
  }
}
int main() {
  red.push_back(1);
  int a1, a2;
  cin >> n >> q;
  int sq = sqrt(q), sz = 0;
  for (int i = 0; i < 30; ++i) par[1][i] = 1;
  for (int i = 1; i < n; ++i) {
    cin >> a1 >> a2;
    ne[a1].push_back(a2);
    ne[a2].push_back(a1);
  }
  dfs(1);
  for (int i = 0; i < MAXN; ++i) D.push_back(INF);
  for (int i = 0; i < q; ++i) {
    cin >> a1 >> a2;
    if (a1 == 1) {
      red.push_back(a2);
      if (red.size() % sq == 0) {
        upd();
        sz = red.size();
      }
      continue;
    }
    int mn = D[a2];
    for (int k = sz; k < red.size(); ++k) mn = min(mn, dis(red[k], a2));
    cout << mn << '\n';
  }
  return 0;
}
