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;
long long n, m, i, j, rt, sz[100005], dep[100005], fr[100005], lk[100005],
    par[100005], vis[100005];
vector<long long> e[100005];
set<long long> son[100005];
priority_queue<pair<long long, long long> > pq[100005];
priority_queue<pair<long long, long long> > pq2;
vector<long long> v[100005];
void dfs1(long long x, long long fa) {
  sz[x] = 1;
  bool f = 1;
  int i;
  for (i = 0; i < e[x].size(); i++) {
    if (e[x][i] == fa) continue;
    dfs1(e[x][i], x);
    sz[x] += sz[e[x][i]];
    if (sz[e[x][i]] > n / 2) f = 0;
  }
  if (n - sz[x] > n / 2) f = 0;
  if (f) rt = x;
}
void dfs2(long long x, long long fa, long long f) {
  fr[x] = f;
  sz[x] = 1;
  int i;
  for (i = 0; i < e[x].size(); i++) {
    if (e[x][i] == fa) continue;
    par[e[x][i]] = x;
    son[x].insert(e[x][i]);
    dep[e[x][i]] = dep[x] + 1;
    dfs2(e[x][i], x, fa ? f : e[x][i]);
    sz[x] += sz[e[x][i]];
  }
  if (f && !son[x].empty()) pq[f].push(make_pair(dep[x], x));
}
void dfs3(long long x, long long fa, long long f) {
  if (f && !lk[x]) v[f].push_back(x);
  int i;
  for (i = 0; i < e[x].size(); i++) {
    if (e[x][i] == fa) continue;
    dfs3(e[x][i], x, fa ? f : e[x][i]);
  }
}
void solve(int t) {
  while (!pq[t].empty() && (pq[t].top().first > m || vis[pq[t].top().second])) {
    pq[t].pop();
  }
  assert(!pq[t].empty());
  int x = pq[t].top().second;
  pq[t].pop();
  if (son[x].size() >= 2) {
    int t1 = *son[x].begin();
    son[x].erase(son[x].begin());
    int t2 = *son[x].begin();
    son[x].erase(son[x].begin());
    lk[t1] = t2;
    lk[t2] = t1;
    if (!son[x].empty()) pq[t].push(make_pair(dep[x], x));
  } else {
    int t1 = *son[x].begin();
    son[x].erase(son[x].begin());
    lk[t1] = x;
    lk[x] = t1;
    son[par[x]].erase(x);
    if (son[par[x]].empty()) vis[par[x]] = 1;
  }
  sz[t] -= 2;
  m -= dep[x];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m;
  for ((i) = 1; (i) <= (n - 1); (i)++) {
    long long x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(rt, 0, 0);
  for ((i) = 1; (i) <= (n); (i)++) {
    m -= dep[i];
  }
  m = -m;
  if (m < 0 || m % 2 != 0) {
    puts("NO");
    return 0;
  }
  m /= 2;
  for (__typeof((son[rt]).begin()) it = (son[rt]).begin();
       it != (son[rt]).end(); ++it) {
    pq2.push(make_pair(sz[*it], *it));
  }
  while (m) {
    if (pq2.empty()) {
      puts("NO");
      return 0;
    }
    int t = pq2.top().second;
    pq2.pop();
    if (sz[t] < 2) {
      puts("NO");
      return 0;
    }
    solve(t);
    if (sz[t]) pq2.push(make_pair(sz[t], t));
  }
  dfs3(rt, 0, 0);
  while (pq2.size() > 1) {
    int t1 = pq2.top().second;
    pq2.pop();
    int t2 = pq2.top().second;
    pq2.pop();
    sz[t1]--;
    sz[t2]--;
    lk[v[t1].back()] = v[t2].back();
    lk[v[t2].back()] = v[t1].back();
    v[t1].pop_back();
    v[t2].pop_back();
    if (sz[t1]) pq2.push(make_pair(sz[t1], t1));
    if (sz[t2]) pq2.push(make_pair(sz[t2], t2));
  }
  assert(pq2.size() == 1);
  int t = v[pq2.top().second][0];
  lk[t] = rt;
  lk[rt] = t;
  cout << "YES\n";
  for ((i) = 1; (i) <= (n); (i)++)
    if (i < lk[i]) cout << i << ' ' << lk[i] << endl;
  return 0;
}
