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最新推荐文章于 2024-06-28 14:18:34 发布

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本文介绍了一种计算特定长度及数值范围内非羊毛序列数量的方法。羊毛序列是指包含一段连续子序列，其异或值为0的整数序列。文中提供了一个高效的算法实现，并通过样例解释了计算原理。

A sequence of non-negative integers a1, a2, ..., an of length n is called a wool sequence if and only if there exists two integers l and r(1 ≤ l ≤ r ≤ n) such that $\text{[math]}$ . In other words each wool sequence contains a subsequence of consecutive elements with xor equal to 0.

The expression $\text{[math]}$ means applying the operation of a bitwise xor to numbers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is marked as "^", in Pascal — as "xor".

In this problem you are asked to compute the number of sequences made of n integers from 0 to 2m - 1 that are not a wool sequence. You should print this number modulo 1000000009 (109 + 9).

Input

The only line of input contains two space-separated integers n and m (1 ≤ n, m ≤ 105).

Output

Print the required number of sequences modulo 1000000009 (109 + 9) on the only line of output.

Sample test(s)

input
3 2

output
6

Note

Sequences of length 3 made of integers 0, 1, 2 and 3 that are not a wool sequence are (1, 3, 1), (1, 2, 1), (2, 1, 2), (2, 3, 2), (3, 1, 3) and (3, 2, 3).

从这题学到了一个小技巧：处理一串数的抑或，可以像前缀和那样先预处理出前缀的抑或值，然后再枚举端点就可以得到任意一段的抑或值。然后这题要求任意一段的抑或都不相等，也就是前缀抑或数组b中不能有相等的元素，因为b中每个数的取值范围都是1~2^m-1（根绝题目条件，任何一个数都不能取0），且任意给b赋值，原来的数组a都一定存在一个唯一解，即一一对应关系。这样答案就是排列数A（2^m-1,n）.

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <cmath>
#include <string>
using namespace std;
typedef long long LL;
const int maxn = 3000 + 5;
const int INF = 100000;
const int Mod = 1000000000 + 9;

LL cal(LL x){
    LL ret = 1;
    while(x > 30){
        ret = (((LL)1 << 30) * ret) % Mod;
        x -= 30;
    }
    ret = (((LL)1 << x) * ret) % Mod;
    return ret;
}

int main(){
    LL n,m;
    while(cin >> n >> m){
        LL ans = 1;
        LL pow = cal(m);
        for(LL i = 1;i <= n;i++){
            ans = (ans * (pow - i)) % Mod;
        }
        cout << ans << endl;
    }

    return 0;
}