#pragma GCC optimize("O3")
#include<bits/stdc++.h> 
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
template<class T> using V=vector<T>; 
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
const ll inf=(1e18);
// const ll mod=998244353;
const ll mod=1000000007;
const vector<int> dy={-1,0,1,0},dx={0,-1,0,1};
struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<<fixed<<setprecision(15);}} __init;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
template<class T>void debag(const vector<T> &a){cerr<<"debag :";for(auto v:a)cerr<<v<<" ";cerr<<"\n";}
template<class T>void print(const vector<T> &a){for(auto v:a)cout<<v<<" ";cout<<"\n";}
struct mint{
using ull=unsigned long long int;
ull v;
mint(ll vv=0){s(vv%mod+mod);}
mint& s(ull vv){
v=vv<mod?vv:vv-mod;
return *this;
}
//オーバーロード
mint operator-()const{return mint()-*this;}//mint型にキャスト
mint&operator+=(const mint&val){return s(v+val.v);}
mint&operator-=(const mint&val){return s(v+mod-val.v);}
mint&operator*=(const mint&val){
v=ull(v)*val.v%mod;
return *this;
}
mint&operator/=(const mint&val){return *this*=val.inv();}
mint operator+(const mint&val){return mint(*this)+=val;}
mint operator-(const mint&val){return mint(*this)-=val;}
mint operator*(const mint&val){return mint(*this)*=val;}
mint operator/(const mint&val){return mint(*this)/=val;}
mint pow(ll n)const{
mint res(1),x(*this);
while(n){
if(n&1)res*=x;
x*=x;
n>>=1ll;
}
return res;
}
mint inv()const{return pow(mod-2);}
//拡張ユークリッドの互除法
/* mint inv()const{
int x,y;
int g=extgcd(v,mod,x,y);
assert(g==1);
if(x<0)x+=mod;
return mint(x);
}*/
friend ostream& operator<<(ostream&os,const mint&val){
return os<<val.v;
}//出力
bool operator<(const mint&val)const{return v<val.v;}
bool operator==(const mint&val)const{return v==val.v;}
bool operator>(const mint&val)const{return v>val.v;}
};
void solve(){
    ll a,n;
    cin>>a>>n;
    ll m=ll(sqrt(double(a*n+100)));
     while(m*m<a*n)m++;
    while(m*m>a*n)m--;
    mint ans=mint(2)*mint(m).pow(3)+mint(3)*mint(m).pow(2)-mint(5)*mint(m);
    ans/=mint(6);
    mint sump=mint(0),sumq=mint(0);
    ll p=m/a,q=m%a;
    for(ll i=1;i<=a;i++){
        if(i<=q)sumq+=mint((i*i-1)%a);
        sump+=mint((i*i-1)%a);
    }
    ans-=mint(p)*sump+sumq;
    ans/=mint(a);
    ans=mint(n)*mint(m)-ans;
    cout<<ans<<"\n";
}
int main(){
    int t;
    cin>>t;
    while(t--)solve();
}