Problem Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.



"The second problem is, given an positive integer N, we define an equation like this:

  N=a[1]+a[2]+a[3]+...+a[m];

  a[i]>0,1<=m<=N;

My question is how many different equations you can find for a given N.

For example, assume N is 4, we can find:

  4 = 4;

  4 = 3 + 1;

  4 = 2 + 2;

  4 = 2 + 1 + 1;

  4 = 1 + 1 + 1 + 1;

so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

 

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

 

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

 

Sample Input

4

10

20

 

Sample Output

5

42

627

/*#include<stdio.h>

#include<string.h>

int dp[1000][1000],n;

int main()

{

	while(scanf("%d",&n)!=EOF)

	{

		memset(dp,0,sizeof(dp));

		for(int i=1;i<1000;i++)

		dp[i][1]=dp[1][i]=1;

		for(int i=2;i<=n;i++)

		{

			for(int j=1;j<=n;j++)

			{

				if(i>j) dp[i][j]=dp[i-j][j]+dp[i][j-1];

				else if(i==j) dp[i][j]=dp[i][j-1]+1;

				else dp[i][j]=dp[i][i];

			}

		}

		printf("%d\n",dp[n][n]);

	}

	return 0;

}*/

#include<stdio.h>

#include<string.h>

int dp[1200];

int main()

{

	int n;

	while(scanf("%d",&n)!=EOF)

	{

		memset(dp,0,sizeof(dp));

		dp[0]=1;

		for(int i=1;i<=n;i++)

		for(int j=i;j<=n;j++)

		{

			dp[j]+=dp[j-i];

		}

		printf("%d\n",dp[n]);

	}

}