Codeforces Round #411 (Div. 2)(A,B,C,D 四水题)

Tavak and Seyyed are good friends. Seyyed is very funny and he told Tavak to solve the following problem instead of longest-path.

You are given l and r. For all integers from l to r, inclusive, we wrote down all of their integer divisors except 1. Find the integer that we wrote down the maximum number of times.

Solve the problem to show that it's not a NP problem.

Input

The first line contains two integers l and r (2 ≤ l ≤ r ≤ 10⁹).

Output

Print single integer, the integer that appears maximum number of times in the divisors.

If there are multiple answers, print any of them.

Examples

Input

19 29

Output

Input

3 6

Output

Note

Definition of a divisor: https://www.mathsisfun.com/definitions/divisor-of-an-integer-.html

The first example: from 19 to 29 these numbers are divisible by 2: {20, 22, 24, 26, 28}.

The second example: from 3 to 6 these numbers are divisible by 3: {3, 6}.

题目链接：http://codeforces.com/contest/805/problem/A

分析：直接判断l==r就输出l，否则输出2就可以了！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int l,r;
 6     scanf("%d%d",&l,&r);
 7     if(l==r)
 8         cout<<l<<endl;
 9     else cout<<2<<endl;
10     return 0;
11 }

In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.

He is too selfish, so for a given n he wants to obtain a string of n characters, each of which is either 'a', 'b' or 'c', with no palindromes of length 3 appearing in the string as a substring. For example, the strings "abc" and "abca" suit him, while the string "aba" doesn't. He also want the number of letters 'c' in his string to be as little as possible.

Input

The first line contains single integer n (1 ≤ n ≤ 2·10⁵) — the length of the string.

Output

Print the string that satisfies all the constraints.

If there are multiple answers, print any of them.

Examples

Input

Output

aa

Input

Output

bba

Note

A palindrome is a sequence of characters which reads the same backward and forward.

题目链接：http://codeforces.com/contest/805/problem/B

分析：直接输出aabb.........这组样例即可，要多少个输出多少个，比如n=5 ，aabba，n=6，aabbaa。。。。。。

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     while(scanf("%d",&n)!=EOF)
 7     {
 8         for(int i=1;i<=n;++i)
 9         {
10             cout<<"a";
11             ++i;
12             if(i<=n)
13             cout<<"a";
14             ++i;
15             if(i<=n)
16             cout<<"b";
17             ++i;
18             if(i<=n)
19             cout<<"b";
20         }
21         cout<<endl;
22     }
23     return 0;
24 }

A few years ago Sajjad left his school and register to another one due to security reasons. Now he wishes to find Amir, one of his schoolmates and good friends.

There are n schools numerated from 1 to n. One can travel between each pair of them, to do so, he needs to buy a ticket. The ticker between schools i and j costs $\text{[math]}$ and can be used multiple times. Help Sajjad to find the minimum cost he needs to pay for tickets to visit all schools. He can start and finish in any school.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of schools.

Output

Print single integer: the minimum cost of tickets needed to visit all schools.

Examples

Input

Output

Input

Output

Note

In the first example we can buy a ticket between the schools that costs $\text{[math]}$ .

题目链接：http://codeforces.com/contest/805/problem/C

分析：就解释一下第二组数据怎么来的，即为答案！

1－＞10 ，10－＞2， 2－＞9， 9－＞3， 3－＞8 ，8－＞４，４－＞7，7－＞５，5－＞6

(1+10)%11==0;

(10+2)%11==1 ;

(2+9) %11==0;

(9+3)%11==1;

(3+8)%11==0;

(8+4)%11==1;

(4+7)%11==0;

(7+5)%11==1;

(5+6)%11==0;

上式求和sum即为4.。。。。。遍历一遍求值，即可

在探索的过程中，似乎发现了一个公式：sum=(n-1)/2;

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     while(scanf("%d",&n)!=EOF)
 7     {
 8         int sum=n-1;
 9         sum/=2;
10         printf("%d\n",sum);
11     }
12     return 0;
13 }

We have a string of letters 'a' and 'b'. We want to perform some operations on it. On each step we choose one of substrings "ab" in the string and replace it with the string "bba". If we have no "ab" as a substring, our job is done. Print the minimum number of steps we should perform to make our job done modulo 10⁹ + 7.

The string "ab" appears as a substring if there is a letter 'b' right after the letter 'a' somewhere in the string.

Input

The first line contains the initial string consisting of letters 'a' and 'b' only with length from 1 to 10⁶.

Output

Print the minimum number of steps modulo 10⁹ + 7.

Examples

Input

ab

Output

Input

aab

Output

Note

The first example: "ab" → "bba".

The second example: "aab" → "abba" → "bbaba" → "bbbbaa".

题目链接：http://codeforces.com/contest/805/problem/D

分析：用t去记录b得数量，遇到'b'，t++，遇到'a'应该是先加上答案t的值，然后t的值再翻倍！

例如：

aab 遇到a变成 abba 遇到a变成bbbbaa b都是翻倍的增加的

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e9+7;
 4 int main()
 5 {
 6     string s;
 7     cin>>s;
 8     int ans=0;
 9     int t=0;//b的数量
10     for(int i=s.length()-1;i>=0;i--)
11     {
12         if(s[i]=='b')
13             t=(t+1)%N;
14         else
15         {
16             ans=(ans+t)%N;
17             t=(t*2)%N;
18         }
19     }
20     cout<<ans<<endl;
21     return 0;
22 }