You are given an integer k and a tree T with n nodes (n is even).

Let dist(u, v) be the number of edges on the shortest path from node u to node v in T.

Let us define a undirected weighted complete graph G = (V, E) as following: 

  * V = \\{x ∣ 1 ≤ x ≤ n \} i.e. the set of integers from 1 to n
  * E = \{(u, v, w) ∣ 1 ≤ u, v ≤ n, u ≠ v, w = dist(u, v) \} i.e. there is an edge between every pair of distinct nodes, the weight being the distance between their respective nodes in T 



Your task is simple, find a perfect matching in G with total edge weight k (1 ≤ k ≤ n^2).

Input

The first line of input contains two integers n, k (2 ≤ n ≤ 100 000, n is even, 1 ≤ k ≤ n^2): number of nodes and the total edge weight of the perfect matching you need to find.

The i-th of the following n - 1 lines contains two integers v_i, u_i (1 ≤ v_i, u_i ≤ n) denoting an edge between v_i and u_i in T. It is guaranteed that the given graph is a tree. 

Output

If there are no matchings that satisfy the above condition, output "NO" (without quotes) on a single line. 

Otherwise, you should output "YES" (without quotes) on the first line of output.

You should then output n/2 lines, the i-th line containing p_i, q_i (1 ≤ p_i, q_i ≤ n): the i-th pair of the matching.

Examples

Input


4 2
1 2
2 3
3 4


Output


YES
2 1
3 4


Input


4 4
1 2
2 3
3 4


Output


YES
3 1
2 4

Note

A tree is a connected acyclic undirected graph.

A matching is set of pairwise non-adjacent edges, none of which are loops; that is, no two edges share a common vertex. 

A perfect matching is a matching which matches all vertices of the graph; that is, every vertex of the graph is incident to exactly one edge of the matching.
#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 20;
long long n, k;
vector<long long> g[N];
long long sz[N], h[N];
long long cen;
set<pair<long long, long long> > second, st[N];
set<long long> child[N];
long long par[N];
long long match[N];
void pre_dfs1(long long v, long long par = -1) {
  sz[v] = 1;
  for (long long u : g[v]) {
    if (u == par) continue;
    pre_dfs1(u, v);
    sz[v] += sz[u];
  }
}
long long find_centroid(long long v, long long par = -1) {
  for (long long u : g[v]) {
    if (u == par) continue;
    if (sz[u] * 2 > n) return find_centroid(u, v);
  }
  return v;
}
void pre_dfs2(long long v, long long cc = -1) {
  sz[v] = 1;
  for (long long u : g[v]) {
    if (u == par[v]) continue;
    par[u] = v;
    child[v].insert(u);
    h[u] = h[v] + 1;
    long long ncc = (cc == -1 ? u : cc);
    pre_dfs2(u, ncc);
    sz[v] += sz[u];
  }
  if (cc != -1) st[cc].insert({h[v], v});
  if (v == cc) second.insert({sz[v], v});
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 0, u, v; i < n - 1; i++) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  pre_dfs1(0);
  cen = find_centroid(0);
  par[cen] = -1;
  h[cen] = 0;
  pre_dfs2(cen);
  for (long long i = 0; i < n; i++) k -= h[i];
  k = -k;
  if (k < 0 || k % 2 == 1) return cout << "NO\n", 0;
  k /= 2;
  for (long long i = 0; i < n; i++) match[i] = -1;
  while (k) {
    pair<long long, long long> p = *second.rbegin();
    second.erase(p);
    if (p.first == 1) return cout << "NO\n", 0;
    long long m = p.second;
    pair<long long, long long> q = *st[m].rbegin();
    long long v = q.second, pp = par[v];
    if (k > h[pp]) {
      child[pp].erase(v);
      if (((long long)child[pp].size()) > 0) {
        long long u = *child[pp].begin();
        child[pp].erase(u);
        match[v] = u;
        match[u] = v;
        st[m].erase({h[v], v});
        st[m].erase({h[u], u});
      } else {
        match[v] = pp;
        match[pp] = v;
        child[par[pp]].erase(pp);
        st[m].erase({h[v], v});
        st[m].erase({h[pp], pp});
      }
      k -= h[pp];
      p.first -= 2;
      if (p.first) second.insert(p);
    } else {
      for (pair<long long, long long> p : st[m]) {
        long long v = p.second;
        if (h[v] == k && ((long long)child[v].size()) > 0) {
          long long u = *child[v].begin();
          match[u] = v;
          match[v] = u;
          st[m].erase({h[v], v});
          st[m].erase({h[u], u});
          break;
        }
      }
      p.first -= 2;
      if (p.first) second.insert(p);
      k = 0;
      break;
    }
  }
  while (((long long)second.size()) > 1) {
    pair<long long, long long> p1 = *second.rbegin();
    second.erase(p1);
    pair<long long, long long> p2 = *second.rbegin();
    second.erase(p2);
    long long m1 = p1.second, m2 = p2.second;
    long long u = st[m1].begin()->second;
    long long v = st[m2].begin()->second;
    match[u] = v;
    match[v] = u;
    st[m1].erase({h[u], u});
    st[m2].erase({h[v], v});
    p1.first--;
    p2.first--;
    if (p1.first) second.insert(p1);
    if (p2.first) second.insert(p2);
  }
  {
    long long m = second.begin()->second;
    long long v = st[m].begin()->second;
    match[cen] = v;
    match[v] = cen;
  }
  cout << "YES\n";
  for (long long i = 0; i < n; i++)
    if (match[i] > i) cout << i + 1 << ' ' << match[i] + 1 << '\n';
}
