Graph Theory Class
原理不会板子抄的。

// n^0.75/log 求1~n的质数和
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 10000010;ll n,mod;
int prime[N],id1[N],id2[N],vis[N],cnt,m;
ll g[N],s[N],a[N],T;
int ID(ll x){return x<=T?id1[x]:id2[n/x];}
ll calc(ll x){return x*(x+1)/2-1;}
ll f(ll x){return x;}
ll init(ll n)
{T=sqrt(n+0.5);for(int i=2;i<=T;i++) {if(!vis[i])prime[++cnt]=i,s[cnt]=s[cnt-1]+i;for(int j=1;j<=cnt&&i*prime[j]<=T;j++) {vis[i*prime[j]]=1;if(i%prime[j]==0)break;}}for (ll l=1;l<=n;l=n/(n/l)+1) {a[++m]=n/l;if(a[m]<=T)id1[a[m]]=m; else id2[n/a[m]]=m;g[m]=calc(a[m]);}for(int i=1;i<=cnt;i++)for(int j=1;j<=m&&(ll)prime[i]*prime[i]<=a[j];j++)g[j]=g[j]-(ll)prime[i]*(g[ID(a[j]/prime[i])]-s[i-1]);
}
ll solve(ll x)// solve(x)表示求1~x的质数和
{if(x<=1)return x;return n=x,init(n),g[ID(n)];
}
void clear()// 多组数据清空
{memset(g,0,sizeof(g));memset(a,0,sizeof(a));memset(s,0,sizeof(s));memset(prime,0,sizeof(prime));memset(id1,0,sizeof(id1));memset(id2,0,sizeof(id2));memset(vis,0,sizeof(vis));cnt=m=0;
}
int main() {int Tc;scanf("%d",&Tc);while(Tc--){clear();scanf("%lld%lld",&n,&mod);ll ans=0;if(n&1) ans=(n-1)/2%mod*(n+4)%mod;else ans=(n+4)/2%mod*(n-1)%mod;ans=(ans+solve(n+1)-2)%mod;ans=(ans+mod)%mod;printf("%lld\n",ans);}return 0;
}