Problem:
Score : 700 points

Problem Statement

Determine if there exists a sequence obtained by permuting 1,2,...,N that satisfies the following conditions:

The length of its longest increasing subsequence is A.

The length of its longest decreasing subsequence is B.

If it exists, construct one such sequence.

Notes

A subsequence of a sequence P is a sequence that can be obtained by extracting some of the elements in P without changing the order.

A longest increasing subsequence of a sequence P is a sequence with the maximum length among the subsequences of P that are monotonically increasing.

Similarly, a longest decreasing subsequence of a sequence P is a sequence with the maximum length among the subsequences of P that are monotonically decreasing.

Constraints

1 \leq N,A,B \leq 3\times 10^5

All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

If there are no sequences that satisfy the conditions, print -1.

Otherwise, print N integers. The i-th integer should be the i-th element of the sequence that you constructed.

Sample Input 1

5 3 2

Sample Output 1

2 4 1 5 3

One longest increasing subsequence of this sequence is {2,4,5}, and one longest decreasing subsequence of it is {4,3}.

Sample Input 2

7 7 1

Sample Output 2

1 2 3 4 5 6 7

Sample Input 3

300000 300000 300000

Sample Output 3

-1

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main() {
	long long int  n,a,b,memo;
	scanf("%lld%lld%lld", &n,&a,&b);
	memo = 0;
	if (b == 1 && a == n) {
		for (int i = 1; i <= n; i++) {
			printf("%lld ", i);
			memo--;
          
		}
		return 0;
	}
	else if (b == n && a == 1) {
		for (int i = 1; i <= n; i++) {
			printf("%lld ", memo);
			memo--;
          
		}
      	return 0;
	}
	else if (a == 1 || b == 1 || a == n || b == n||a+b>n+1||n/2>=a&&n/2>=b) { printf("-1"); return 0; }

	if (a >= b) {
		for (int i = n - a + 1; i <= n; i++) {
			printf("%lld ", i);

		}
		for (int i = b - 1; i <= n - a; i++) {
			printf("%lld ", i);
		}
		for (int i = b - 2; i >= 1; i--){
			printf("%lld ", i);
		}
	}
	else {
		for (int i = b; i >= 1; i--) {
			printf("%lld ", i);
		}
		for (int i = n-a+2; i >= b+1; i--) {
			printf("%lld ", i);
		}
		for (int i = n - a+3; i <= n; i++) {
			printf("%lld ", i);
		}
	}

}