#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int T;
cin >> T;
while (T--) {
int N, M;
cin >> N >> M;
// Let P[] be the prefix sum of A[]. In particular,
// P[0] = 0, P[1] = A[1], P[2] = A[1] + A[2], ...
//
// P[] can be arbitary since A[] can be arbitary.
// The condition is:
// P[i + N] - P[i] < 0 -> P[i + N] < P[i]
// P[i + M] - P[i] > 0 -> P[i + M] > P[i].
// We draw an edge x -> y if P[x] < P[y]. Then, if there
// is a strongly connected component, that particular size
// is invalid.
//
// Now, note that if size >= N + M, there is a cycle. Proof:
// WLOG, assume N <= M. Then, P[i] < P[i + M] and P[i] < P[i - N].
// We start from node = 0, then node += M, then decrease by N until
// node = (node_prv) % N. We do this again, and again. Note that
// the maximum index is <= N + M, and eventually we will arrive back
// at 0, since we form a cycle at modulo N.
//
// We can binary search for the answer, checking whether the graph
// is a DAG or not. Time complexity: O((N + M) log (N + M)). This
// yields 76 points.
//
// Can we get a tight bound? Consider the process, of +M and -N.
// We can actually simulate this process pretty easily without DFS.
// Time complexity: O(N + M).
vector<int> P(N + M + 1);
vector<vector<int>> adj(N + M + 1);
for (int i = 0; i <= N + M; i++) {
if (i <= N) adj[i].emplace_back(i + M);
if (i >= N) adj[i].emplace_back(i - N);
}
vector<int> topo;
vector<int> vis(N + M + 1);
vector<int> pos_in_topo(N + M + 1);
const auto Calc = [&](int sz) -> bool {
topo.clear();
fill(begin(vis), end(vis), 0);
const auto Dfs = [&](const auto &self, int u) -> void {
vis[u] = 1;
for (auto v : adj[u]) if (v <= sz && !vis[v]) self(self, v);
topo.emplace_back(u);
};
for (int i = 0; i <= sz; i++) if (!vis[i]) {
Dfs(Dfs, i);
}
reverse(begin(topo), end(topo));
for (int i = 0; i <= sz; i++) {
pos_in_topo[topo[i]] = i;
}
for (int i = 0; i <= sz; i++) {
for (auto j : adj[i]) if (j <= sz) {
if (pos_in_topo[j] < pos_in_topo[i]) {
return false;
}
}
}
for (int i = 0; i <= sz; i++) P[topo[i]] = i;
for (int i = sz; i >= 0; i--) P[i] -= P[0];
return true;
};
int ans = 0;
int node = 0;
fill(begin(vis), end(vis), 0);
while (!vis[node]) {
vis[node] = 1;
ans = max(ans, node);
if (node >= N) {
node -= N;
} else {
node += M;
}
}
ans--;
assert(node == 0);
Calc(ans);
cout << ans << '\n';
for (int i = 1; i <= ans; i++) {
cout << (P[i] - P[i - 1]) << " \n"[i == ans];
}
}
return 0;
}
