#include<bits/stdc++.h>
using namespace std;
//#pragma GCC optimize("Ofast")

#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=(n)-1;i>=0;i--)
#define perl(i,r,l) for(ll i=r-1;i>=l;i--)
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
#define vtpl(x,y,z) vector<tuple<x,y,z>>
#define rev(x) reverse(x);
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
using vpl=vector<pl>;
using vvpl=vector<vpl>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[9]={1,0,-1,0,1,1,-1,-1,0};
const ll dx[9]={0,1,0,-1,1,-1,1,-1,0};
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

//https://atcoder.jp/contests/abc230/submissions/44985883
vector<tuple<ll,ll,ll>> quotient_floors(ll n){
    //floor(n/i)=xを満たすx,i∈[l,r)をこの順に返す。lの昇順、つまりxの降順なことに注意。
    vector<tuple<ll,ll,ll>> ret;
    for(ll l=1;l<=n;){
        ll x=n/l;
        ll r=n/x+1;
        ret.emplace_back(x,l,r);
        l=r;
    }
    return ret;
}

int main(){
    ll p,q;cin >> p >> q;
    auto fs=quotient_floors(p);
    ll m=fs.size();
    vl cum(m+1);
    rep(i,m){
        auto [x,l,r]=fs[i];
        cum[i+1]=cum[i]+(r-l)*p-x*(r+l-1)*(r-l)/2;
    }
    vl left(m);rep(i,m)left[i]=get<1>(fs[i]);
    //rep(i,m)cout << left[i] <<" ";cout << endl;
    while(q--){
        ll ans=0;
        ll L,R;cin >> L >>R;R++;
        ll idx=upper_bound(all(left),R)-left.begin();idx--;
        //cout << idx << endl;
        ans+=cum[idx];
        if(idx!=m){
            auto [x,l,r]=fs[idx];
            r=min(r,R);
            ans+=(r-l)*p-x*(r+l-1)*(r-l)/2;
        }
        idx=upper_bound(all(left),L)-left.begin();idx--;
        //cout << idx << endl;
        ans-=cum[idx];
        if(idx!=m){
            auto [x,l,r]=fs[idx];
            r=min(r,L);
            ans-=(r-l)*p-x*(r+l-1)*(r-l)/2;
        }
        ans+=p*max(R-1-p,0LL);
        ans-=p*max(L-1-p,0LL);
        cout << ans << endl;
    }
}