You are given three integers n, d and k.

Your task is to construct an undirected tree on n vertices with diameter d and degree of each vertex at most k, or say that it is impossible.

An undirected tree is a connected undirected graph with n - 1 edges.

Diameter of a tree is the maximum length of a simple path (a path in which each vertex appears at most once) between all pairs of vertices of this tree.

Degree of a vertex is the number of edges incident to this vertex (i.e. for a vertex u it is the number of edges (u, v) that belong to the tree, where v is any other vertex of a tree).

Input

The first line of the input contains three integers n, d and k (1 ≤ n, d, k ≤ 4 ⋅ 10^5).

Output

If there is no tree satisfying the conditions above, print only one word "NO" (without quotes).

Otherwise in the first line print "YES" (without quotes), and then print n - 1 lines describing edges of a tree satisfying the conditions above. Vertices of the tree must be numbered from 1 to n. You can print edges and vertices connected by an edge in any order. If there are multiple answers, print any of them.1

Examples

Input

6 3 3


Output

YES
3 1
4 1
1 2
5 2
2 6


Input

6 2 3


Output

NO


Input

10 4 3


Output

YES
2 9
2 10
10 3
3 1
6 10
8 2
4 3
5 6
6 7


Input

8 5 3


Output

YES
2 5
7 2
3 7
3 1
1 6
8 7
4 3
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, d, k;
  cin >> n >> d >> k;
  vector<int> w(n + 1, 0), first(n + 1, 0), v(n + 1, -1);
  queue<int> q;
  q.push(1);
  int maxf = 0, maxs = 0, idf = 0;
  for (int c = 2; c <= n;) {
    if (q.empty()) {
      cout << "NO";
      return 0;
    }
    int x = q.front();
    q.pop();
    for (int j = 0; j < k - 1 + (x == 1) && c <= n; j++, c++) {
      q.push(c);
      w[c] = w[x] + 1;
      v[c] = x;
      first[c] = first[x];
      if (c == 2) first[c] = 1;
      if (first[c]) {
        maxf = w[c];
        idf = c;
      } else
        maxs = w[c];
      if (n - c + maxs + maxf == d && c != n && k != 1) {
        c += 1;
        for (; n - c >= 0; c++) {
          w[c] = w[idf] + 1;
          v[c] = idf;
          idf = c;
          maxf = w[c];
        }
      }
    }
  }
  if (maxf + maxs == d) {
    cout << "YES\n";
    for (int i = 2; i <= n; i++) {
      cout << i << " " << v[i] << "\n";
    }
  } else {
    cout << "NO";
  }
}
