#pragma GCC optimize ("O3")
#include "bits/stdc++.h"
using namespace std;

#define rep(i, b, e) for(int i = (b); i <= (e); i++)
#define per(i, b, e) for(int i = (e); i >= (b); i--)
#define FOR(i, b, e) rep(i, b, (e) - 1)
#define SZ(x) int(x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define st first
#define nd second
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

auto &operator<<(auto &o, pair<auto, auto> p) {
    return o << "(" << p.st << ", " << p.nd << ")";
}

auto operator<<(auto &o, auto x)->decltype(end(x), o) {
    o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
    return o << "}";
}

#ifdef LOCAL
#define deb(x...) cerr << "[" #x << "]: ", [](auto...$) { ((cerr << $ << "; "),...) << endl; }(x)
#else
#define deb(...)
#endif

const int N = 18 * 18;
using B = bitset<N>;

B lans;
int ans;

void cliq(vector<B> &eds, B P = ~B(), B X={}, B R={}) {
    if(!P.any()) {
        int res = R.count();
        if(res > ans) {
            ans = res;
            lans = R;
        } 
        return;
    }
    auto q = (P | X)._Find_first();
    auto cands = P & ~eds[q];
    if((int)(R.count() + cands.count()) <= ans) return;
    FOR(i, 0, SZ(eds)) if(cands[i]) {
        R[i] = 1;
        cliq(eds, P & eds[i], X & eds[i], R);
        R[i] = P[i] = 0; X[i] = 1;
    }
}

void solve() {
    int n;
    double r;
    cin >> n >> r;
    map<pii, int> ind;
    map<int, pii> rind;
    int nr = 0;
    int bound = ceil(r);
    rep(i, bound, n-bound) rep(j, bound, n-bound) {
        rind[nr] = {i, j};
        ind[{i, j}] = nr++;
    }
    vector<B> g(nr);
    set<int> dists;
    rep(i, bound, n-bound) rep(j, bound, n-bound) rep(k, bound, n-bound) rep(l, bound, n-bound) {
        int dst = (i - k) * (i - k) + (j - l) * (j - l);
        if(dst >= 4 * r * r - 1e-3) g[ind[{i, j}]][ind[{k, l}]] = 1;
        dists.insert(dst);
    }
    deb(SZ(dists));
    deb(dists);
    cliq(g);
    cout << ans << '\n';
    FOR(i, 0, nr) if(lans[i]) cout << rind[i].st << ' ' << rind[i].nd << '\n';
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int tt = 1;
    // cin >> tt;

    
    FOR(te, 0, tt) solve();
    return 0;
}