/* Counting sort : linear time sorting algorithm that sorts the array in O(n + k)
                   where values are in range [1, k]

   What if the elements are in [1, n^2] -> O(n^2) complexity worse than any comparison sorting algorithm
  Solution: RADIX sort
  Algorithm
   Do the following for each digit i ranging from least to the most significant one
			sort the input array based on the ith digit using counting sort

  Coplexity O(n*#digits) where #digits ~ log(maxx)
  Useful when the data has a large number of dupliucates and the value range is relatively small
*/

#include <bits/stdc++.h>
using namespace std;
const int nmax = 5e5+2;
FILE *fin  = freopen("algsort.in", "r", stdin);
FILE *fout = freopen("algsort.out", "w", stdout);

int n, a[nmax];

void countingsort(int exp) {
	int count[10] = {0};
	int output[nmax];

	for (int i = 0; i < n; ++ i)
		count[(a[i]/exp)%10] ++;

	// compute the position of the last element with digit i
	for (int i = 1; i < 10; ++ i)
		count[i] += count[i-1];

	// compute the output array
	for(int i = n-1; i >= 0; -- i) {
        count[(a[i]/exp)%10] --;
		output[count[(a[i]/exp)%10]] = a[i];
	}
	for (int i = 0; i < n; ++ i)
		a[i] = output[i];
}

void radixsort() {
	int maxx = -INT_MAX;
	for (int i = 0; i < n; ++ i)
		maxx = max(a[i], maxx);

	// apply counting sort for every digit
    for (int exp = 1; maxx/exp > 0; exp *= 10)
		countingsort(exp);
}

int main() {
	cin >> n;
	for (int i = 0 ; i < n; ++ i)
		cin >> a[i];
	radixsort();
	for (int i = 0; i < n; ++ i)
		cout << a[i] << " ";
	return 0;
}