#include #include #include #include #include #include #include using namespace std; typedef long long ll; typedef long double ld; typedef pair pr; const double pi=acos(-1); #define rep(i,a,n) for(int i=a;i<=n;i++) #define per(i,n,a) for(int i=n;i>=a;i--) #define Rep(i,u) for(int i=head[u];i;i=Next[i]) #define clr(a) memset(a,0,sizeof a) #define pb push_back #define mp make_pair ld eps=1e-9; ll pp=998244353; ll mo(ll a,ll pp){if(a>=0 && a>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;} ll read(){ ll ans=0; char last=' ',ch=getchar(); while(ch<'0' || ch>'9')last=ch,ch=getchar(); while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar(); if(last=='-')ans=-ans; return ans; } //head #define N 1100000 int prime[N],rec[N],u[N],num=0; bool vis[N]; ll s[N][2][2]; int n,m; void solved(){ n=read(),m=read(); if(n=1;i--) if(u[i]!=0){ for(int j=i*2;j<=m;j+=i) rep(k1,0,1) rep(k2,0,1) s[i][k1][k2]-=s[j][k1][k2]; int cnt; if(u[i]==1)cnt=0; else cnt=1; rep(k1,0,1) rep(k2,0,1){ //printf("%d %d %d : %d\n",i,k1,k2,(int)s[i][k1][k2]); if((k1+k2+cnt)%2)ans1+=s[i][k1][k2]; else ans0+=s[i][k1][k2]; } } cout<