An n?×?n table a is defined as follows:
The first row and the first column contain ones, that is: ai,?1?=?a1,?i?=?1 for all i?=?1,?2,?...,?n.
Each of the remaining numbers in the table is equal to the sum of the number above it and the number to the left of it. In other words, the remaining elements are defined by the formula ai,?j?=?ai?-?1,?j?+?ai,?j?-?1.
These conditions define all the values in the table.
You are given a number n. You need to determine the maximum value in the n?×?n table defined by the rules above.
Input
The only line of input contains a positive integer n (1?≤?n?≤?10) — the number of rows and columns of the table.
Output
Print a single line containing a positive integer m — the maximum value in the table.
Examples
Input
1
Output
1
Input
5
Output
70
Note
In the second test the rows of the table look as follows:
{1,?1,?1,?1,?1},?
{1,?2,?3,?4,?5},?
{1,?3,?6,?10,?15},?
{1,?4,?10,?20,?35},?
{1,?5,?15,?35,?70}.

定義一個(gè)二維數(shù)組，讓其第一行和第一列為1，其余由其加得，輸出最后一個(gè)數(shù)即可。

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int a[11][11];
    for (int i = 0; i < n; i++)
    {
        a[0][i] = 1;
    }
    for (int i = 1; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (j == 0)
            {
                a[i][j] = 1;
            }
            else
                a[i][j] = a[i][j - 1] + a[i - 1][j];
        }
    }
    cout << a[n-1][n-1];
    return 0;
}