#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
const int mxm=2e5+5;
const ll N=1e10+5;
bool isprime[mxm];
ll prime[mxm],w[mxm],id1[mxm],id2[mxm];
ll num_pri=0;
ll m=0;
ll sump[mxm];
__int128 g[mxm];
__int128 cal(ll x){
return (__int128)x*(x+1)/2-1;
}
ll sqr;
ll n,mod;
ll id3(ll x){
return x<=sqr?id1[x]:id2[n/x];
}
void el(){
sqr=sqrt(N);
isprime[1]=1;
for(ll i=2;i<=sqr;++i){
if(!isprime[i]) prime[++num_pri]=i,sump[num_pri]=sump[num_pri-1]+i;
for(ll j=1;j<=num_pri&&i*prime[j]<=sqr;++j){
isprime[i*prime[j]]=1;
if(i%prime[j]==0) break;
}
}
}
void init(){
for(ll i=1;i<=n;i=n/(n/i)+1){
w[++m]=n/i;
if(w[m]<=sqr){
id1[w[m]]=m;
}
else{
id2[n/w[m]]=m;
}
g[m]=cal(w[m]);
}
for(ll i = 1;i<=num_pri;++i){
for(ll j=1; j<=m && 1ll*prime[i]*prime[i] <= w[j]; ++j){
g[j]=g[j] - (__int128)prime[i] * (g[id3(w[j] / prime[i])] - sump[i - 1]);
}
}
}
int main(){
el();
int t;
scanf("%d",&t);
while(t--){
scanf("%lld%lld",&n,&mod);
++n;
if(n==2){
puts("0");
continue;
}
m=0;
init();
ll x = n + 3;
ll y = n - 3 + 1;
if ((n + 3) & 1) y /= 2;
else x /= 2;
__int128 answer = (x % mod) * (y % mod) % mod;
answer += g[id3(n)] - 2;
answer %= mod;
printf("%lld\n", (ll)answer );
}
return 0;
}