codeforces A. Vitya in the Countryside

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本文介绍了一个有趣的编程挑战，即根据连续几天的月相变化来预测接下来一天的月相是上升还是下降，或是无法确定。通过分析给定的月相数据序列，参与者需要编写程序来判断月相的变化趋势。

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Every summer Vitya comes to visit his grandmother in the countryside. This summer, he got a huge wart. Every grandma knows that one should treat warts when the moon goes down. Thus, Vitya has to catch the moment when the moon is down.

Moon cycle lasts 30 days. The size of the visible part of the moon (in Vitya's units) for each day is 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12, 13, 14, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, and then cycle repeats, thus after the second 1 again goes 0.

As there is no internet in the countryside, Vitya has been watching the moon for n consecutive days and for each of these days he wrote down the size of the visible part of the moon. Help him find out whether the moon will be up or down next day, or this cannot be determined by the data he has.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 92) — the number of consecutive days Vitya was watching the size of the visible part of the moon.

The second line contains n integers ai (0 ≤ ai ≤ 15) — Vitya's records.

It's guaranteed that the input data is consistent.

Output

If Vitya can be sure that the size of visible part of the moon on day n + 1 will be less than the size of the visible part on day n, then print "DOWN" at the only line of the output. If he might be sure that the size of the visible part will increase, then print "UP". If it's impossible to determine what exactly will happen with the moon, print -1.

Examples

input
5
3 4 5 6 7

output
UP

input
7
12 13 14 15 14 13 12

output
DOWN

input
1
8

output
-1

Note

In the first sample, the size of the moon on the next day will be equal to 8, thus the answer is "UP".

In the second sample, the size of the moon on the next day will be 11, thus the answer is "DOWN".

In the third sample, there is no way to determine whether the size of the moon on the next day will be 7 or 9, thus the answer is -1.

这个题的坑点是n=1 和a[i]=15 a[i]=0;

代码如下：

#include <stdio.h>

int main()
{
	int n,a[100],i;
	
	while(~scanf("%d",&n))
	{
		if(n==1)
		{
			scanf("%d",&a[0]);
			
			if(a[0]==15)
				printf("DOWN\n");
			else if(a[0]==0)
				printf("UP\n");
			else
				printf("-1\n");
		}
		else
		{
			for(i=0;i<n;i++)
				scanf("%d",&a[i]);
			if(a[n-1]==15)
				printf("DOWN\n");
			else if(a[n-1]==0)
			{
				printf("UP\n");
			}
			else
			{
				if(a[n-2] < a[n-1])
				{
					printf("UP\n");
				}
				else if(a[n-2] > a[n-1])
				{
					printf("DOWN\n");
				}
			}
		}
	}
	
	return 0;
}