Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b >>= 1;
  }
  return res;
}
long long modInv(long long a) { return binpow(a, MOD - 2); }
bool sortcol(const vector<long long>& v1, const vector<long long>& v2) {
  return v1[1] < v2[1];
}
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int pct(int x) { return __builtin_popcount(x); }
int bits(int x) { return 31 - __builtin_clz(x); }
int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }
int fstTrue(function<bool(int)> first, int lo, int hi) {
  hi++;
  assert(lo <= hi);
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    first(mid) ? hi = mid : lo = mid + 1;
  }
  return lo;
}
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<long long unsigned> distribution(0, 10);
const int mxN = 2e5;
int q = 1, n, a, b, c;
long long k, sz[mxN], depth[mxN], paritySum, subtreeSum, par[mxN];
vector<int> adj[mxN];
set<pair<int, int> > subSz;
vector<set<int> > children[mxN];
vector<pair<int, int> > ans;
set<int> child[mxN];
void init_subtrees_and_parents(int x, int d, int root) {
  while (children[root].size() <= d) {
    set<int> second;
    children[root].push_back(second);
  }
  children[root][d].insert(x);
  for (auto& i : adj[x]) {
    if (depth[i] > depth[x]) {
      par[i] = x;
      init_subtrees_and_parents(i, d + 1, root);
    }
  }
}
int init_size_and_depth(int v, int p = -1) {
  if (p != -1) depth[v] = depth[p] + 1;
  sz[v] = 1;
  for (auto& x : adj[v]) {
    if (x != p) {
      sz[v] += init_size_and_depth(x, v);
    }
  }
  return sz[v];
}
int find_centroid(int v, int p, int n) {
  for (auto& x : adj[v]) {
    if (x != p && sz[x] > n / 2) {
      return find_centroid(x, v, n);
    }
  }
  return v;
}
void solve() {
  cin >> n >> k;
  for (int i = (0); i < (n - 1); i++) {
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  init_size_and_depth(0);
  c = find_centroid(0, -1, sz[0]);
  fill(depth, depth + n, 0);
  init_size_and_depth(c);
  for (int i = (0); i < (n); i++) {
    for (auto& j : adj[i]) {
      if (depth[j] > depth[i]) {
        child[i].insert(j);
      }
    }
  }
  for (int i = (0); i < (n); i++) {
    if (i != c) {
      paritySum += sz[i] % 2;
      subtreeSum += sz[i];
    }
  }
  if (k % 2 != paritySum % 2 || k < paritySum || k > subtreeSum)
    cout << "NO\n";
  else {
    cout << "YES\n";
    for (auto& i : adj[c]) subSz.insert({sz[i], i});
    for (auto& i : adj[c]) init_subtrees_and_parents(i, 0, i);
    while (subtreeSum > k) {
      pair<int, int> p = *subSz.rbegin();
      int u = *children[p.second][children[p.second].size() - 1].rbegin();
      while (k > subtreeSum - 2 * (depth[u] - 1)) {
        u = par[u];
      }
      subtreeSum -= 2 * (depth[u] - 1);
      if (child[par[u]].size() == 1) {
        child[par[u]].erase(u);
        child[par[par[u]]].erase(par[u]);
        children[p.second][depth[u] - 1].erase(u);
        children[p.second][depth[u] - 2].erase(par[u]);
        ans.push_back({u, par[u]});
      } else {
        int x = par[u];
        u = *child[x].rbegin();
        child[x].erase(u);
        int v = *child[x].rbegin();
        child[x].erase(v);
        children[p.second][depth[x]].erase(u);
        children[p.second][depth[x]].erase(v);
        ans.push_back({u, v});
      }
      while (!children[p.second][children[p.second].size() - 1].size() &&
             children[p.second].size()) {
        children[p.second].pop_back();
      }
      subSz.erase({p.first, p.second});
      subSz.insert({p.first - 2, p.second});
    }
    while (1) {
      if (subSz.size() <= 1) break;
      pair<int, int> p1 = *subSz.rbegin();
      subSz.erase({p1.first, p1.second});
      pair<int, int> p2 = *subSz.rbegin();
      subSz.erase({p2.first, p2.second});
      if (!p1.first) break;
      if (!p2.first) {
        ans.push_back({p1.second, c});
        break;
      }
      while (!children[p1.second][children[p1.second].size() - 1].size() &&
             children[p1.second].size())
        children[p1.second].pop_back();
      while (!children[p2.second][children[p2.second].size() - 1].size() &&
             children[p2.second].size())
        children[p2.second].pop_back();
      int u = *children[p1.second][children[p1.second].size() - 1].rbegin();
      int v = *children[p2.second][children[p2.second].size() - 1].rbegin();
      children[p1.second][depth[u] - 1].erase(u);
      children[p2.second][depth[v] - 1].erase(v);
      while (!children[p1.second][children[p1.second].size() - 1].size() &&
             children[p1.second].size())
        children[p1.second].pop_back();
      while (!children[p2.second][children[p2.second].size() - 1].size() &&
             children[p2.second].size())
        children[p2.second].pop_back();
      ans.push_back({u, v});
      subSz.insert({p1.first - 1, p1.second});
      subSz.insert({p2.first - 1, p2.second});
    }
    if (subSz.size() == 1 && (*subSz.rbegin()).first)
      ans.push_back({(*subSz.rbegin()).second, c});
    for (auto& i : ans) cout << i.first + 1 << " " << i.second + 1 << "\n";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  while (q--) {
    solve();
  }
}