#include<bits/stdc++.h>
using namespace std;
#define int long long
#define lowbit(x) (x & (-x))
#define lc p << 1
#define rc p << 1 | 1
typedef long long ll;
typedef pair<int,int> pii;
mt19937 rnd(time(0));
const int N = 1e6 + 7;
const int mod = 1e9 + 7;
vector<pii> v[205];
void solve() {
    int l , r; cin >> l >> r;
    int ni = 0,nx = 0,now = 1,ans;
    for(int i = 0;i < 24;i++) if((1 << i) & l) ni = i;
    for(int i = 0;i < 24;i++) if((1 << i) & r) nx = i;
    int st = -1;
    for(int i = nx;i >= 0;i--) {
        if(((1 << i) & l && (1 << i) & r) || (!((1 << i) & l) && !((1 << i) & r))) continue; //上下相同
        else {st = i;break;}
    }
    if(st == -1) {
        for(int i = nx;i >= 0;i--) {
            v[now].push_back({now + 1,(bool)((1 << i) & r)});
            now++;
        }
        ans = now;
    }
    else {
        for(int i = nx;i >= 0;i--) {
            v[now].push_back({now + 1,(bool)((1 << i) & r)});
            now++;
        }
        int lst = now,l1 = 2,r1 = now - 1;
        int l2,r2;
        if(l != 1) {
            now++;  //变成第二个开始的点
            l2 = now,r2 = now;
            v[1].push_back({now,(bool)((1 << ni) & l)});
            for(int i = ni - 1;i > 0;i--) { //第二个点开始
                v[now].push_back({now + 1,(bool)((1 << i) & l)});
                now++;
            }
            r2 = now;
            v[r2].push_back({lst,(bool)(1 & l)});
        }
        else {
            v[1].push_back({lst,1});
            l2 = 0,r2 = -1; //后续找主线最高不进循环
        }
        int mx = 0;
        if(ni != nx) {
            for(int i = r1,j = 0;i >= l1;i--,j++) { 
                if(v[i][0].second == 1) { //找每个点的需要的长，直接连上
                    if(j == 0) v[i].push_back({lst,0});
                    else v[i].push_back({now + j,0});
                    mx = max(mx,now + j);
                }
            }
            for(int i = r2,j = 0;i >= l2;i--,j++) {
                if(v[i][0].second == 0) { //同上
                    if(j == 0) v[i].push_back({lst,1});
                    else v[i].push_back({now + j,1});
                    mx = max(mx,now + j);
                }
            }
        }
        else { //如果位数相等，只能在不等之后看
            for(int i = r1,j = 0;i >= l1 && j < st;i--,j++) { 
                if(v[i][0].second == 1) { //找每个点的需要的长，直接连上
                    if(j == 0) v[i].push_back({lst,0});
                    else v[i].push_back({now + j,0});
                    mx = max(mx,now + j);
                }
            }
            for(int i = r2,j = 0;i >= l2 && j < st;i--,j++) {
                if(v[i][0].second == 0) { //同上
                    if(j == 0) v[i].push_back({lst,1});
                    else v[i].push_back({now + j,1});
                    mx = max(mx,now + j);
                }
            }
        }
        for(int i = ni + 1;i < nx;i++) {
            v[1].push_back({now + i,1}); //找1的最高点
            mx = max(mx,now + i);
        }
        for(int i = mx;i > now;i--) { //把最高点到now所有都铺上
            if(i == now + 1) {
                v[i].push_back({lst,0});
                v[i].push_back({lst,1});
                break;
            }
            v[i].push_back({i - 1,1});
            v[i].push_back({i - 1,0});
        }
        ans = max(mx,now); //如果没主线直接就用当前最大即可
    }
    cout << ans << "\n";
    for(int i = 1;i <= ans;i++) {
        cout << v[i].size() << " ";
        for(auto [x,y] : v[i]) cout << x << " " << y << " ";
        cout << "\n";
    }
}
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    // cin >> t;
    while(t--) solve();
    return 0;
}