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 int maxn = 100010;
int n, rt, tim, sz[maxn], tid[maxn], dep[maxn], vis[maxn], fa[maxn];
long long K, lb, rb;
vector<int> rem, G[maxn];
set<pair<int, int>> Q, lca[maxn], leaf[maxn];
int main() {
  scanf("%d %lld", &n, &K);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    G[u].push_back(v), G[v].push_back(u);
  }
  function<void(int, int)> dfs1 = [&](int v, int fa) {
    int mx = 0;
    sz[v] = 1;
    for (int u : G[v])
      if (u ^ fa) {
        dfs1(u, v), sz[v] += sz[u];
        mx = max(mx, sz[u]);
        lb += sz[u] & 1, rb += min(sz[u], n - sz[u]);
      }
    if (max(mx, n - sz[v]) << 1 <= n) rt = v;
  };
  dfs1(1, 0);
  if (K < lb || K > rb || (K & 1) ^ (lb & 1)) puts("NO"), exit(0);
  puts("YES");
  function<void(int, int)> dfs2 = [&](int v, int bel) {
    dep[v] = dep[fa[v]] + 1, tid[v] = ++tim, sz[v] = 1;
    if (G[v].size() == 1)
      leaf[bel].emplace(tid[v], v);
    else
      lca[bel].emplace(dep[v], v);
    for (int u : G[v])
      if (u ^ fa[v]) {
        fa[u] = v, dfs2(u, bel), sz[v] += sz[u];
      }
  };
  for (int v : G[rt]) {
    fa[v] = rt, dfs2(v, v), Q.emplace(sz[v], v);
  }
  while (rb > K) {
    auto sub = *Q.rbegin();
    int w = sub.second;
    Q.erase(sub);
    if (lca[w].empty()) continue;
    auto chk = [&](int v) {
      if (!lca[w].count({dep[v], v})) return;
      auto p = leaf[w].lower_bound({tid[v], 0});
      if (p == leaf[w].end() || p->first >= tid[v] + sz[v])
        lca[w].erase({dep[v], v}), leaf[w].emplace(tid[v], v);
    };
    auto p = --lca[w].lower_bound({((rb - K) >> 1) + 1, 0});
    int z = p->second;
    rb -= p->first << 1;
    int u = (p = leaf[w].lower_bound({tid[z], 0}))->second, v;
    leaf[w].erase(p), p = leaf[w].lower_bound({tid[z] + sz[z], 0});
    if (p == leaf[w].begin() || (--p)->first < tid[z] || K == rb)
      v = z, lca[w].erase({dep[z], z});
    else
      v = p->second, leaf[w].erase(p);
    printf("%d %d\n", u, v), vis[u] = vis[v] = 1;
    chk(fa[u]), chk(fa[v]);
    sub.first -= 2, Q.emplace(sub);
  }
  function<void(int)> dfs3 = [&](int v) {
    if (!vis[v]) rem.push_back(v);
    for (int u : G[v])
      if (u ^ fa[v]) dfs3(u);
  };
  dfs3(rt);
  for (int i = 0; i < rem.size() / 2; i++) {
    printf("%d %d\n", rem[i], rem[i + rem.size() / 2]);
  }
  return 0;
}
