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;
int n, sz[100010], head[100010], o = 0;
pair<int, int> res[100010];
long long k, mn = 0, mx = 0;
struct edge {
  int to, link;
} e[200010];
void add_edge(int u, int v) {
  e[++o].to = v, e[o].link = head[u], head[u] = o;
  e[++o].to = u, e[o].link = head[v], head[v] = o;
}
void dfs(int u, int pre) {
  sz[u] = 1;
  for (int i = head[u]; i; i = e[i].link) {
    if (e[i].to == pre) continue;
    dfs(e[i].to, u);
    sz[u] += sz[e[i].to];
  }
}
list<int> work(int u, int pre) {
  list<int> tmp;
  tmp.push_back(u);
  for (int i = head[u]; i; i = e[i].link) {
    if (e[i].to == pre) continue;
    tmp.splice(tmp.end(), work(e[i].to, u));
  }
  int size = (tmp.size() - sz[u]) >> 1;
  for (int i = 0; i < size; i++) {
    res[i].first = tmp.front();
    res[size - i - 1].second = tmp.back();
    tmp.pop_front();
    tmp.pop_back();
  }
  for (int i = 0; i < size; i++) printf("%d %d\n", res[i].first, res[i].second);
  return tmp;
}
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 1, u, v; i < n; i++) scanf("%d%d", &u, &v), add_edge(u, v);
  dfs(1, 0);
  for (int i = 1; i <= n; i++)
    mx += (sz[i] = min(sz[i], n - sz[i])), mn += (sz[i] & 1);
  if (k < mn || k > mx || (mx - k) % 2 != 0) return puts("NO"), 0;
  int l = 0, r = n >> 1, mid;
  long long ans = -1;
  while (l <= r) {
    mid = (l + r) >> 1;
    long long tmp = 0;
    for (int i = 2; i <= n; i++) {
      if (sz[i] <= mid)
        tmp += sz[i];
      else
        tmp += mid - ((sz[i] - mid) & 1);
    }
    if (tmp >= k) {
      r = mid - 1;
      ans = tmp;
    } else
      l = mid + 1;
  }
  for (int i = 2; i <= n; i++) {
    if (sz[i] > l) sz[i] = l - ((sz[i] - l) & 1);
    if (sz[i] == l && ans > k) sz[i] -= 2, ans -= 2;
  }
  puts("YES");
  work(1, 0);
}
