Educational Codeforces Round 32 ABC

A. Local Extrema

You are given an array a. Some element of this array a_i is a local minimum iff it is strictly less than both of its neighbours (that is, a_i < a_i - 1 and a_i < a_i + 1). Also the element can be called local maximum iff it is strictly greater than its neighbours (that is, a_i > a_i - 1 and a_i > a_i + 1). Since a₁ and a_n have only one neighbour each, they are neither local minima nor local maxima.

An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.

Input

The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array a.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000) — the elements of array a.

Output

Print the number of local extrema in the given array.

Examples

Input

3
1 2 3

Output

Input

4
1 5 2 5

Output

求极点的数量

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int a[1010];
 4 int main() {
 5     int n;
 6     cin >> n;
 7     for(int i = 1; i <= n; i ++) cin >> a[i];
 8     int ans = 0;
 9     for(int i = 2; i < n; i ++) {
10         if(a[i] > a[i-1] && a[i] > a[i+1]) ans++;
11         if(a[i] < a[i-1] && a[i] < a[i+1]) ans++;
12     }
13     printf("%d
",ans);
14     return 0;
15 }

Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell (0, 0). The robot can process commands. There are four types of commands it can perform:

U — move from the cell (x, y) to (x, y + 1);
D — move from (x, y) to (x, y - 1);
L — move from (x, y) to (x - 1, y);
R — move from (x, y) to (x + 1, y).

Ivan entered a sequence of n commands, and the robot processed it. After this sequence the robot ended up in the starting cell (0, 0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!

Input

The first line contains one number n — the length of sequence of commands entered by Ivan (1 ≤ n ≤ 100).

The second line contains the sequence itself — a string consisting of n characters. Each character can be U, D, L or R.

Output

Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.

Examples

Input

4
LDUR

Output

Input

5
RRRUU

Output

Input

6
LLRRRR

Output

4

求长度最长的命令  会使的回到原点。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 char str[110];
 4 int main() {
 5     int n, a, b, c, d;
 6     a = b = c = d = 0;
 7     cin >> n >> str;
 8     for(int i = 0; i < n; i ++) {
 9         if(str[i] == 'U') a++;
10         else if(str[i] == 'D') b++;
11         else if(str[i] == 'L') c++;
12         else if(str[i] == 'R') d++;
13     }
14     printf("%d
",min(a,b)*2+min(c,d)*2);
15     return 0;
16 }

You are given a string s consisting of lowercase Latin letters. Character c is called k-dominant iff each substring of s with length at least k contains this character c.

You have to find minimum k such that there exists at least one k-dominant character.

Input

The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 100000).

Output

Print one number — the minimum value of k such that there exists at least one k-dominant character.

Examples

Input

abacaba

Output

Input

zzzzz

Output

Input

abcde

Output

在s字符串中，求每个个字符的最大间距的最小值。

00Z000中'z'的最大间距算3

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 100010;
 4 char str[N];
 5 int sum[N];
 6 bool vis[27];
 7 int main() {
 8     cin >> str;
 9     int len = strlen(str);
10     int k = len;
11     for(int i = 0; str[i]; i ++) {
12         sum[i] = str[i] - 'a';
13         vis[sum[i]] = true;
14     }
15     for(int i = 0; i < 26; i ++) {
16         if(vis[i]) {
17             int x = -1, tmp = 0, MAX = -1;
18             for(int j = 0; j < len; j ++) {
19                 if(sum[j] != i) tmp ++;
20                 else {
21                     if(tmp + 1 > MAX) MAX = tmp + 1;
22                     tmp = 0;
23                 }
24             }
25             if(tmp + 1 > MAX) MAX = tmp + 1;
26             if(MAX < k) k = MAX;
27         }
28     }
29     printf("%d
",k);
30     return 0;
31 }