Let us consider sets of positive integers less than or equal to n. Note that all elements of a set are different. Also note that the order of elements doesn‘t matter, that is, both {3, 5, 9} and {5, 9, 3} mean the same set.

Specifying the number of set elements and their sum to be k and s, respectively, sets satisfying the conditions are limited. When n = 9, k= 3 and s = 23, {6, 8, 9} is the only such set. There may be more than one such set, in general, however. When n = 9, k = 3 and s = 22, both {5, 8, 9} and {6, 7, 9} are possible.

You have to write a program that calculates the number of the sets that satisfy the given conditions.

Input

The input consists of multiple datasets. The number of datasets does not exceed 100.

Each of the datasets has three integers n, k and s in one line, separated by a space. You may assume 1 ≤ n ≤ 20, 1 ≤ k ≤ 10 and 1 ≤ s ≤ 155.

The end of the input is indicated by a line containing three zeros.

Output

The output for each dataset should be a line containing a single integer that gives the number of the sets that satisfy the conditions. No other characters should appear in the output.

You can assume that the number of sets does not exceed 2³¹ - 1.

Sample Input

9 3 23
9 3 22
10 3 28
16 10 107
20 8 102
20 10 105
20 10 155
3 4 3
4 2 11
0 0 0

Output for the Sample Input

1
2
0
20
1542
5448
1
0
0


这种题目，一看就知道思路，就是枚举法来做，可是有那么多种情况，N！种，怎么都超时了，可是我们也要注意到一种题目的标志，就是状态压缩，对于这类题目，n一般都是20，而这个题就是20，而且完全符合状态压缩，虽然这个题有dp 的解法，可是状态压缩已经足以解决了

#include<map>
#include<set>
#include<stack>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f
using namespace std;

int sum,k,n,a[25];

int get(int state)
{
    int cut=0;
    while(state)
    {
        a[cut++]=state%2;
        state=state/2;
    }
    return cut;
}

bool find(int s)
{
    int cut=0;
    int ans=0;
    for (int i=0;i<s;i++)
    {
        if (a[i]==1)
        {
            cut++;
            ans+=i+1;
        }
    }
    if (cut==k && ans==sum) return true;
    else return false;
}

int main()
{
    while(scanf("%d%d%d",&n,&k,&sum)!=EOF && n && k && sum)
    {
        int s=1<<n;
        int ans=0;
        for (int i=0;i<s;i++)
        {
            int b=get(i);
            if (find(b)) ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}

作者 chensunrise