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 N = 2e5 + 5;
const int mod = 1e9 + 7;
long long m, n, t, k, a[N], b[N], tong, rt, ans, deep, h[N], u, v, sz[N], siz,
    mx[N], par[N], P[N][20];
vector<long long> adj[N], adt[N];
void findprt(long long u, long long p) {
  sz[u] = 1;
  mx[u] = 0;
  for (long long v : adj[u]) {
    if (v == p || b[v] == -1) continue;
    findprt(v, u);
    sz[u] += sz[v];
    mx[u] = max(mx[u], sz[v]);
  }
  mx[u] = max(mx[u], siz - sz[u]);
  if (!rt || mx[u] < mx[rt]) rt = u;
}
void dfs(long long u, long long p) {
  for (int i = 1; i <= 18; i++) P[u][i] = P[P[u][i - 1]][i - 1];
  for (long long v : adj[u]) {
    if (v == p) continue;
    h[v] = h[u] + 1;
    P[v][0] = u;
    dfs(v, u);
  }
}
void centriod(long long root, long long len) {
  b[root] = -1;
  deep = max(deep, len);
  long long all = siz, k = sz[root];
  for (long long v : adj[root]) {
    if (b[v] == -1) continue;
    siz = sz[v] > k ? all - k : sz[v];
    rt = 0;
    findprt(v, 0);
    par[rt] = root;
    centriod(rt, len + 1);
  }
}
long long lca(long long u, long long v) {
  if (h[u] < h[v]) swap(u, v);
  long long log = log2(h[u]);
  for (int i = log; i >= 0; i--)
    if (h[u] >= h[v] + (1 << i)) u = P[u][i];
  if (u == v) return u;
  for (int i = log; i >= 0; i--) {
    if (P[u][i] != P[v][i] && P[u][i] != 0) {
      u = P[u][i];
      v = P[v][i];
    }
  }
  return P[u][0];
}
long long dis(long long u, long long v) {
  long long z = lca(u, v);
  return h[u] + h[v] - 2 * h[z];
}
void up(long long u, long long v) {
  if (u == 0) return;
  a[u] = min(a[u], dis(u, v));
  up(par[u], v);
}
long long get(long long u, long long v) {
  if (u == 0) return n + 1;
  return min(a[u] + dis(u, v), get(par[u], v));
}
void sol() {
  cin >> n >> m;
  for (int i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) a[i] = n + 1;
  h[1] = 1;
  dfs(1, 1);
  siz = n;
  findprt(1, 1);
  centriod(rt, 1);
  a[1] = 0;
  up(1, 1);
  for (int i = 1; i <= m; i++) {
    long long x, y;
    cin >> x >> y;
    if (x == 1) {
      a[y] = 0;
      up(y, y);
    } else {
      cout << get(y, y) << '\n';
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (fopen("coupons"
            ".in",
            "r")) {
    freopen(
        "coupons"
        ".in",
        "r", stdin);
    freopen(
        "coupons"
        ".out",
        "w", stdout);
  }
  sol();
}
