题目：给n个不超过m的素数，求xor和=0的方案数，FWT变换裸题。
题目

2关于F逆元的公式： $inv(2)=(F+1)>>1$
证：$ [(F+1)>>1]*2 \pmod F=F+1 \pmod F =1$

代码：

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll F=1000000007LL;
ll iv2=F+1>>1; //Õâ¸ö¼ÆËã˼·ºÃÆÀ 
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
void fwt(int*a,int n){
    int i,j,k,x;
    for(k=2;k<=n;k<<=1){
        for(i=0;i<n;i+=k){
            for(j=i;j<i+(k>>1);j++){
                x=a[j]+a[j+(k>>1)];if(x>=F) x-=F;
                a[j]=a[j]-a[j+(k>>1)];if(a[j]<0) a[j]+=F;
                a[j+(k>>1)]=x;
            }
        }
    }
}
void twf(int*a,int n){
    int i,j,k,x;
    for(k=n;k>=2;k>>=1){
        for(i=0;i<n;i+=k){
            for(j=i;j<i+(k>>1);j++){
                x=a[j]+a[j+(k>>1)];
                a[j+(k>>1)]=(int)(1LL*(a[j+(k>>1)]-a[j]+F)*iv2%F);
                a[j]=(int)(1LL*x*iv2%F);
            }
        }
    }
}
ll pow2(ll a,ll b){
    if (!b) return 1%F;
    ll p=pow2(a,b/2);
    p=mul(p,p);
    if (b&1) p=mul(p,a);
    return p;
}
#define MAXN (50000<<3)
int b[MAXN],a[MAXN];
ll n,m;
int main()
{
//  freopen("bzoj4589.in","r",stdin);
//  freopen(".out","w",stdout);

    n=50000;
    MEM(b) b[1]=b[0]=1;
    Fork(i,2,n) if (!b[i]) {
        for(int j=2;i*j<=n;j++)
            b[i*j]=1;
    }   
    while(cin>>n>>m) {
        int len=1;
        while(len<=m) len<<=1;
        Rep(i,len) a[i]=(!b[i])&&(i<=m);
        fwt(a,len);
        Rep(i,len) a[i]=pow2(a[i],n);
        twf(a,len);
        printf("%d\n",a[0]);
    }

    return 0;
}