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 = 100000 + 10;
struct Edge {
  int v, next;
  Edge() {}
  Edge(int v, int next) : v(v), next(next) {}
} edges[maxn << 1];
int head[maxn], nEdge;
int d[maxn], mind[maxn], pa[maxn][22];
vector<int> rednode;
void AddEdges(int u, int v) {
  edges[++nEdge] = Edge(v, head[u]);
  head[u] = nEdge;
  edges[++nEdge] = Edge(u, head[v]);
  head[v] = nEdge;
}
void Init() {
  memset(head, 0xff, sizeof(head));
  memset(pa, 0, sizeof(pa));
  memset(mind, 0x3f, sizeof(mind));
  d[1] = 0;
  nEdge = -1;
}
void dfs(int u, int fa) {
  for (int k = head[u]; k != -1; k = edges[k].next) {
    int v = edges[k].v;
    if (v == fa) continue;
    pa[v][0] = u;
    d[v] = d[u] + 1;
    dfs(v, u);
  }
}
void lca_Init(int n) {
  for (int i = 1; i <= 20; ++i)
    for (int j = 1; j <= n; ++j) pa[j][i] = pa[pa[j][i - 1]][i - 1];
}
int Lca(int x, int y) {
  if (x == y) return x;
  if (d[x] > d[y]) swap(x, y);
  for (int i = 20; i >= 0; --i)
    if (d[pa[y][i]] > d[x]) y = pa[y][i];
  if (pa[y][0] == x) return x;
  if (d[y] > d[x]) y = pa[y][0];
  for (int i = 20; i >= 0; --i) {
    if (pa[y][i] != pa[x][i]) {
      y = pa[y][i];
      x = pa[x][i];
    }
  }
  return pa[x][0];
}
void bfs() {
  queue<int> q;
  int sz = rednode.size();
  for (int i = 0; i < sz; ++i) {
    int v = rednode[i];
    q.push(v);
    mind[v] = 0;
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int k = head[u]; k != -1; k = edges[k].next) {
      int v = edges[k].v;
      if (mind[v] > mind[u] + 1) {
        mind[v] = mind[u] + 1;
        q.push(v);
      }
    }
  }
  rednode.clear();
}
int main() {
  int n, m, u, v;
  scanf("%d%d", &n, &m);
  Init();
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    AddEdges(u, v);
  }
  dfs(1, -1);
  lca_Init(n);
  rednode.clear();
  rednode.push_back(1);
  bfs();
  int size = sqrt(m) + 1, cnt = 0;
  while (m--) {
    scanf("%d%d", &u, &v);
    if (u == 1) {
      rednode.push_back(v);
      cnt++;
      if (cnt == size) {
        bfs();
        cnt = 0;
      }
    } else {
      int sz = rednode.size();
      int ans = mind[v];
      for (int i = 0; i < sz; ++i) {
        u = rednode[i];
        int anc = Lca(u, v);
        ans = min(ans, d[u] + d[v] - 2 * d[anc]);
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}
