#include <stdio.h>
#include <stdlib.h>
#include <math.h>

/*
 * observations:
 *
 * solutions of any length L starting at b have the sum
 * S = L * (L-1 + 2b ) / 2 
 *
 * This sum must equal the target t, so the base for a given length is
 * t = L * (L-1+2b) / 2
 * b = (2t/L-L+1)/2
 *
 * Since b must be an integer, t % ( (L+1)*L / 2 ) must equal 0
 * 2t % L must be zero and
 * 2t/L - L + 1 must be even
 *
 * With a minimum base of 1, the sum is (L+1)*L / 2, 
 * which must not exceed the target
 * t >= (L+1)*L/2
 * L <= (sqrt(8t+1)-1) / 2
 *
 * Even targets must have solution lengths >= 3
 * even length solutions can start on any number
 * odd length solutions must be centered on an even number
 *
 * Odd targets must have an odd number of odd numbers
 * in the solution
 * they always have an even length solution of t/2 + t/2 + 1
 *
 */
int main(int ac, char *av[]) {
	unsigned target; /* number to decompose */
	unsigned sbase; /* lowest number in solution */
	unsigned slen; /*  N quantity of numbers in solution */
	int oddtarget;
	unsigned maxlen;

	target = atoi(av[1]);

	maxlen = (sqrt(8.0*target+1.0)-1) / 2;
	oddtarget = target % 2;

	if (oddtarget) {
		printf("2 %d .. %d\n", target/2, target/2+1);
	}

	for (slen=3; slen <= maxlen; slen++) {
		if (oddtarget && (slen % 2 == 0 && (slen/2 % 2 == 0))) {
			continue;
		}

		if (2 * target % slen != 0) {
			continue;
		}
		if ( (2 * target / slen - slen + 1) % 2 != 0) {
			continue;
		}
 
		sbase = (2 * target / slen - slen + 1) / 2;
 	
		printf("%d %d .. %d\n", slen, sbase, sbase+slen-1);
	}

	return 0;
}