// binary search the length
// construct a graph of len + 1 vertices
// draw a directed edge from u to v if we require pref[u] - pref[v] < 0
// start indexing pref[x] from source to sink
// as you can see if we traverse the graph in that manner
// the pref[x]s will be numbered in a way s.t. all requirements are fulfilled
// get final array from pref
// complexity O(T * (N + M)log(N + M)), very tight to TL
// now we will prove that the final length is N + M - gcd(N, M) - 1
// first of all, we can show that if len = N + M - gcd(N, M),
// there is a cycle in the set of vertices divisible by gcd(N, M)
// if we remove vertex N + M - gcd(N, M), the cycle will break as (N + M - gcd(N, M)) is divisible by gcd(N, M)
// thus N + M - gcd(N, M) - 1 is max
// complexity O(T * (N + M))
#include <stdio.h>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
#include <numeric>
#include <queue>
#include <assert.h>
#include <map>
#include <set>
#include <limits.h>
using namespace std;
#define ll long long
#define ld long double
const int MX = 200005;
const int LG = (int)log2(MX) + 2;
const ll mod = 1e9 + 7;
const ll inv2 = (mod + 1) / 2;
#define gc getchar//_unlocked //can't for window server
void cin(int &x){
char c = gc(); bool neg = false;
for(; c < '0'||'9' < c; c = gc())
if(c == '-') neg=true;
x = c - '0'; c = gc();
for(; '0' <= c && c <= '9'; c = gc())
x = (x << 1) + (x << 3) + (c - '0');
if(neg) x = -x;
}
int tc, n, m;
pair<bool, vector<int> > check(int len){
vector<int> deg(len + 1, 0);
for(int i = 0; i <= len; i ++){
if(i - n >= 0) deg[i - n] ++;
if(i + m <= len) deg[i + m] ++;
}
vector<int> pref(len + 1, 0);
queue<int> q;
for(int i = 0; i <= len; i ++){
if(deg[i] == 0)
q.push(i);
}
int cnt = 0;
for(; q.size();){
int nw = q.front(); q.pop();
pref[nw] = ++ cnt;
if(nw - n >= 0){
deg[nw - n] --;
if(deg[nw - n] == 0)
q.push(nw - n);
}
if(nw + m <= len){
deg[nw + m] --;
if(deg[nw + m] == 0)
q.push(nw + m);
}
}
bool valid = 1;
for(int i = 0; i <= len; i ++)
if(deg[i] > 0) valid = 0;
vector<int> res(len, 0);
for(int i = 0; i < len; i ++)
res[i] = pref[i + 1] - pref[i];
return make_pair(valid, res);
}
int main(){
cin(tc);
for(; tc --;){
cin(n); cin(m);
int rs = n + m - __gcd(n, m) - 1;
vector<int> ans = check(rs).second;
printf("%d\n", rs);
for(int i = 0; i < rs; i ++)
printf("%d ", ans[i]);
printf("\n");
}
return 0;
}
