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;
constexpr long long mod = 1000000007;
const long long INF = mod * mod;
const long double eps = 1e-12;
const long double pi = acosl(-1.0);
long long mod_pow(long long x, long long n, long long m = mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % m;
    x = x * x % m;
    n >>= 1;
  }
  return res;
}
struct modint {
  long long n;
  modint() : n(0) { ; }
  modint(long long m) : n(m) {
    if (n >= mod)
      n %= mod;
    else if (n < 0)
      n = (n % mod + mod) % mod;
  }
  operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint& a, modint b) {
  a.n += b.n;
  if (a.n >= mod) a.n -= mod;
  return a;
}
modint operator-=(modint& a, modint b) {
  a.n -= b.n;
  if (a.n < 0) a.n += mod;
  return a;
}
modint operator*=(modint& a, modint b) {
  a.n = ((long long)a.n * b.n) % mod;
  return a;
}
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, long long n) {
  if (n == 0) return modint(1);
  modint res = (a * a) ^ (n / 2);
  if (n % 2) res = res * a;
  return res;
}
long long inv(long long a, long long p) {
  return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
const int max_n = 1000;
modint fact[max_n], factinv[max_n];
void init_f() {
  fact[0] = modint(1);
  for (int i = 0; i < max_n - 1; i++) {
    fact[i + 1] = fact[i] * modint(i + 1);
  }
  factinv[max_n - 1] = modint(1) / fact[max_n - 1];
  for (int i = max_n - 2; i >= 0; i--) {
    factinv[i] = factinv[i + 1] * modint(i + 1);
  }
}
modint comb(int a, int b) {
  if (a < 0 || b < 0 || a < b) return 0;
  return fact[a] * factinv[b] * factinv[a - b];
}
int gcd(int a, int b) {
  if (a < b) swap(a, b);
  while (b) {
    int r = a % b;
    a = b;
    b = r;
  }
  return a;
}
int g;
vector<int> G[1 << 17];
int find_g(int n, int id, int fr) {
  int res = 1;
  int ma = 0;
  for (int to : G[id])
    if (to != fr) {
      int nex = find_g(n, to, id);
      ma = max(ma, nex);
      res += nex;
    }
  if (ma <= n / 2 && n - res <= n / 2) g = id;
  return res;
}
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  find_g(n, 0, -1);
  vector<int> par(n);
  vector<int> depth(n);
  long long mi = 0;
  long long ma = 0;
  function<bool(int, int)> dfs = [&](int id, int fr) -> bool {
    bool res = true;
    int c = 0;
    par[id] = fr;
    for (int to : G[id])
      if (to != fr) {
        depth[to] = depth[id] + 1;
        if (dfs(to, id)) c++;
      }
    mi += 2 * (c / 2);
    if (c % 2) {
      res = false;
      mi++;
    }
    ma += depth[id];
    return res;
  };
  dfs(g, -1);
  if (k < mi || ma < k) {
    cout << "NO\n";
    return;
  }
  long long dif = ma - k;
  if (dif % 2) {
    cout << "NO\n";
    return;
  }
  vector<set<pair<int, int>>> st;
  set<pair<int, int>> cur;
  function<void(int, int)> calc = [&](int id, int fr) {
    cur.insert({depth[id], id});
    for (int to : G[id])
      if (to != fr) {
        calc(to, id);
      }
  };
  priority_queue<pair<int, int>> q;
  for (int to : G[g]) {
    cur.clear();
    calc(to, g);
    st.push_back(cur);
    q.push({cur.size(), st.size() - 1});
  }
  vector<set<int>> exis(n);
  for (int i = 0; i < n; i++) {
    if (par[i] >= 0) {
      exis[par[i]].insert(i);
    }
  }
  vector<pair<int, int>> ans;
  while (dif > 0) {
    pair<int, int> p = q.top();
    q.pop();
    if (p.first - 2 > 0) q.push({p.first - 2, p.second});
    int id = p.second;
    int las = (*st[id].rbegin()).second;
    if (2 * (depth[las] - 1) >= dif) {
      pair<int, int> cur = {dif / 2 + 1, -1};
      st[id].insert(cur);
      auto itr = st[id].find(cur);
      itr++;
      int z = (*itr).second;
      st[id].erase(cur);
      ans.push_back({z, par[z]});
      st[id].erase({depth[z], z});
      st[id].erase({depth[par[z]], par[z]});
      break;
    } else {
      pair<int, int> r = *st[id].rbegin();
      int z = r.second;
      int pa = par[z];
      if (exis[pa].size() >= 2) {
        int le = (*exis[pa].begin());
        exis[pa].erase(exis[pa].begin());
        int ri = (*exis[pa].begin());
        exis[pa].erase(exis[pa].begin());
        ans.push_back({le, ri});
        st[id].erase({depth[le], le});
        st[id].erase({depth[ri], ri});
      } else {
        exis[pa].erase(z);
        exis[par[pa]].erase(pa);
        ans.push_back({z, pa});
        st[id].erase({depth[z], z});
        st[id].erase({depth[pa], pa});
      }
      dif -= 2 * (depth[las] - 1);
    }
  }
  while (q.size() >= 2) {
    pair<int, int> pl = q.top();
    q.pop();
    pair<int, int> pr = q.top();
    q.pop();
    if (pl.first - 1 > 0) {
      q.push({pl.first - 1, pl.second});
    }
    if (pr.first - 1 > 0) {
      q.push({pr.first - 1, pr.second});
    }
    int idl = pl.second, idr = pr.second;
    int vl = (*st[idl].rbegin()).second;
    int vr = (*st[idr].rbegin()).second;
    ans.push_back({vl, vr});
    st[idl].erase({depth[vl], vl});
    st[idr].erase({depth[vr], vr});
  }
  pair<int, int> p = q.top();
  int id = p.second;
  id = (*st[id].begin()).second;
  ans.push_back({g, id});
  cout << "YES\n";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i].first + 1 << " " << ans[i].second + 1 << "\n";
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
  return 0;
}