Codeforces Round #373 (Div. 2) A. Vitya in the Countryside —— 基础题

题目链接：http://codeforces.com/contest/719/problem/A

A. Vitya in the Countryside

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题解：

虽然是基础题，但是却因为细节错了好多遍，所以应当吸取教训。

可知：不管n为多少， 当a[n] = 15时，下一个必定为0， 所以DOWN； 当a[n] = 0时， 下一个必定为1， 所以UP。对于剩下的，如果n>=2,就按正常情况判断；否则为-1。

这题用 if else 来选择，写法应该是最简的。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <sstream>
#include <algorithm>
using namespace std;
#define pb push_back
#define mp make_pair
#define ms(a, b)  memset((a), (b), sizeof(a))
#define eps 0.0000001
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 100000+10;

int main()
{
    int n, a[100];
    scanf("%d",&n);

    for(int i = 1; i<=n; i++)
        scanf("%d",&a[i]);
    
    if(a[n]==0)     //特殊情况
        puts("UP");
    else if(a[n]==15)   //特殊情况
        puts("DOWN");
    else if( n>1 && a[n]>a[n-1] )   //正常判断
        puts("UP");
    else if( n>1 && a[n]<a[n-1] )   //正常判断
        puts("DOWN");
    else 
        puts("-1");
}