#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;
static const int MAXN = 100000;
int n, m, p, timer;
vector<pii> adj[MAXN+1]; // (neighbor, edge_id)
int tin[MAXN+1], low[MAXN+1]; // Tarjan arrays
bool isTreeEdge[200000+5];
char ans[200000+5];
pii edges[200000+5];
// 1) Build DFS‐tree, compute tin/low, mark tree‐edges
void dfs1(int u, int pe) {
tin[u] = low[u] = ++timer;
for (auto [v, eid] : adj[u]) {
if (eid == pe) continue;
if (!tin[v]) {
isTreeEdge[eid] = true;
dfs1(v, eid);
low[u] = min(low[u], low[v]);
} else {
low[u] = min(low[u], tin[v]);
}
}
}
// 2) dfs2 returns net flow through node u
// +x means x more paths need to go upward through parent
// –x means x more paths come down from parent into this subtree
int dfs2(int u, int pe, vector<int>& cnt) {
int subtotal = cnt[u];
for (auto [v, eid] : adj[u]) {
if (!isTreeEdge[eid] || eid == pe) continue;
// only traverse DOWN the tree, skipping the reverse link
int childFlow = dfs2(v, eid, cnt);
// decide ans[eid] by childFlow:
// 0 → no path crosses → 'B'
// >0 → flow goes UP (v→u) → orient accordingly
// <0 → flow goes DOWN (u→v)
if (childFlow == 0) {
ans[eid] = 'B';
} else if (childFlow > 0) {
// child→parent
// if edges[eid] == (u,v) in input order: L = u→v, R = v→u
ans[eid] = (edges[eid].first == u ? 'R' : 'L');
} else {
// parent→child
ans[eid] = (edges[eid].first == u ? 'L' : 'R');
}
subtotal += childFlow;
}
return subtotal;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> m;
for(int i = 0; i < m; i++){
int u,v;
cin >> u >> v;
edges[i] = {u,v};
adj[u].emplace_back(v,i);
adj[v].emplace_back(u,i);
}
// 1) Build tree + lowlinks
dfs1(1, -1);
// 2) Prepare a counter of how many query‐paths start/end at each node
vector<int> cnt(n+1, 0);
// Non‐tree edges are always 'B' and contribute +1/–1 to their endpoints
for(int i = 0; i < m; i++){
if(!isTreeEdge[i]){
ans[i] = 'B';
auto [u,v] = edges[i];
cnt[u]++;
cnt[v]--;
}
}
// Read the p query‐pairs (a→b)
cin >> p;
while(p--){
int a,b;
cin >> a >> b;
cnt[a]++;
cnt[b]--;
}
// 3) dfs2 from root 1 — it returns the net flow for its component
dfs2(1, -1, cnt);
// 4) Print
for(int i = 0; i < m; i++){
cout << ans[i];
}
cout << "\n";
return 0;
}
