要点:
1. 不重：同值元素相对位置不变；
2. 不漏：枚举不遗漏;

perm_0参考y总讲解，实际运行速度更快些，并且天然保持字典顺序；
perm_1参考通过交换源数组[]的元素位置来获取全排列，按照字典顺序输出需要重排序；

def perm_0(arr):
    res = []
    n = len(arr)
    undo = [True] * n 
    path = [0] * n

    def dfs(i):
        if i >= n - 1:
            res.append(list(arr))
            return 
        j = 0 
        while j < n:
            if undo[j]:
                undo[j] = False
                path[i] = arr[j]
                dfs(i + 1)
                undo[j] = True 
            while j + 1 < n and arr[j]== arr[j + 1]: j += 1
            j += 1
    dfs(0)
    return res


def perm_1(arr):
    res = []
    n = len(arr)

    def dfs(i):
        if i >= n:
            res.append(list(arr))
            return 
        j = i
        while j < n:
            nojump=True
            val = arr[j]
            for p in range(i, j):
                if arr[p] == val:
                    nojump = False
                    break
            if nojump:
                arr[i], arr[j] = arr[j], arr[i]
                dfs(i + 1)
                arr[i], arr[j] = arr[j], arr[i] 
            j += 1
    dfs(0)
    return sorted(res)


perm = perm_0
perm = perm_1

n = int(input())
arr = sorted(map(int, input().split()))

for cur in  perm_1(arr):
    print(' '.join([str(i) for i in cur]))