White, Black and White Again ）（CodeForces

C. White, Black and White Again

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus is sure that his life fits the description: "first there is a white stripe, then a black one, then a white one again". So, Polycarpus is sure that this rule is going to fulfill during the next n days. Polycarpus knows that he is in for w good events and b not-so-good events. At least one event is going to take place during each day. As each day is unequivocally characterizes as a part of a white or a black stripe, then each day is going to have events of the same type only (ether good or not-so-good).

What is the number of distinct ways this scenario can develop over the next n days if Polycarpus is in for a white stripe (a stripe that has good events only, the stripe's length is at least 1 day), the a black stripe (a stripe that has not-so-good events only, the stripe's length is at least 1 day) and a white stripe again (a stripe that has good events only, the stripe's length is at least 1 day). Each of n days will belong to one of the three stripes only.

Note that even the events of the same type are distinct from each other. Even if some events occur on the same day, they go in some order (there are no simultaneous events).

Write a code that prints the number of possible configurations to sort the events into days. See the samples for clarifications on which scenarios should be considered distinct. Print the answer modulo 1000000009 (109 + 9).

Input

The single line of the input contains integers n, w and b (3 ≤ n ≤ 4000, 2 ≤ w ≤ 4000, 1 ≤ b ≤ 4000) — the number of days, the number of good events and the number of not-so-good events. It is guaranteed that w + b ≥ n.

Output

Print the required number of ways modulo 1000000009 (109 + 9).

Examples

input

3 2 1

output

2

input

4 2 2

output

4

input

3 2 2

output

4

Note

We'll represent the good events by numbers starting from 1 and the not-so-good events — by letters starting from 'a'. Vertical lines separate days.

In the first sample the possible ways are: "1|a|2" and "2|a|1". In the second sample the possible ways are: "1|a|b|2", "2|a|b|1", "1|b|a|2" and "2|b|a|1". In the third sample the possible ways are: "1|ab|2", "2|ab|1", "1|ba|2" and "2|ba|1".

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int mod=1000000009;long long  c[4010][4010];long long je[4010];int main(){     int n,w,b;     scanf("%d%d%d",&n,&w,&b);     int tmp=max(max(n,w),b);     for(int i=0;i<=tmp;i++)     {         c[i][0]=1;     }     for(int i=1;i<=tmp;i++)     {         for(int j=1;j<=i;j++)         {             c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;         }     }     int tmp2=max(w,b);     je[0]=1;     for(int i=1;i<=tmp2;i++)     {         je[i]=(je[i-1]*(long long)i)%mod;     }     long long ans=0;     for(int i=1;i<=b;i++)     {         long long tmp=n-i-1;         if(!tmp)            break;//乘tmp的意思是：从每个空格里插入坏事情，有多少种情况         tmp=(tmp*c[b-1][i-1])%mod;  //c[b-1][i-1】 是从b-1个空格里插入i-1个空格，i是坏事情的天数，所以是i-1个空格         tmp=(tmp*je[b])%mod;  //有b的阶乘种情况         tmp=(tmp*c[w-1][n-1-i])%mod;  //相当于在所剩的天数里插入n-1-i+1-1，这样的空格，因为再其中的一个地方插入了所有的坏事情所以要减一；         tmp=(tmp*je[w])%mod;//有w的阶乘种情况         ans=(ans+tmp)%mod;  //记录答案     }     cout<<ans<<endl;}