//solution : number of combinations <= 9
#include <bits/stdc++.h>
using namespace std;
#define FOR(i, l, r) for(int i = (l); i < (r); ++i)
#define ROF(i, r, l) for(int i = (r) - 1; i >= (l); --i)
#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second
#define all(v) begin(v), end(v)
#define rall(v) rbegin(v), rend(v)
#define pb push_back
#define eb emplace_back
#define sz(v) (int)v.size()
#define sum_of(v) accumulate(all(v), 0ll)
#define compact(v) v.erase(unique(all(v)), end(v))
#define dbg(x) "[" #x " = " << (x) << "]"
template<typename T>
bool minimize(T& a, const T& b){
if(a > b) return a = b, true;
return false;
}
template<typename T>
bool maximize(T& a, const T& b){
if(a < b) return a = b, true;
return false;
}
using ll = long long;
using db = double;
using ld = long double;
using ull = unsigned long long;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vl = vector<ll>;
using vd = vector<db>;
using vpi = vector<pi>;
using vpl = vector<pl>;
void setIO(){
ios_base::sync_with_stdio(0); cin.tie(0);
#ifdef LOCAL
freopen("task.inp", "r", stdin);
freopen("task.out", "w", stdout);
#endif // LOCAL
}
const int MAX = 2e5 + 5;
int N, Q, num;
string S[3], T;
vector<string> candidates;
bool is_equal[9][MAX << 2];
bool same_all[3][9][MAX << 2];
int lazy[9][MAX << 2];
char g(char a, char b){
if(a == b) return a;
if(a != 'J' && b != 'J') return 'J';
if(a != 'O' && b != 'O') return 'O';
if(a != 'I' && b != 'I') return 'I';
assert(false);
}
string f(string& a, string& b){
string c = "";
assert(sz(a) == sz(b));
FOR(i, 0, sz(a)){
c.pb(g(a[i], b[i]));
}
return c;
}
void build(int id, int l, int r){
if(l == r){
FOR(ids, 0, num){
if(candidates[ids][l] == 'J') same_all[0][ids][id] = 1;
if(candidates[ids][l] == 'O') same_all[1][ids][id] = 1;
if(candidates[ids][l] == 'I') same_all[2][ids][id] = 1;
is_equal[ids][id] = (candidates[ids][l] == T[l]);
}
} else{
int mid = l + r >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
FOR(ids, 0, num){
FOR(i, 0, 3) same_all[i][ids][id] = same_all[i][ids][id << 1] & same_all[i][ids][id << 1 | 1];
is_equal[ids][id] = is_equal[ids][id << 1] & is_equal[ids][id << 1 | 1];
}
}
}
bool ok(){
FOR(i, 0, num) if(is_equal[i][1]) return 1;
return 0;
}
void apply(int ids, int id, int v){
is_equal[ids][id] = same_all[v][ids][id];
lazy[ids][id] = v;
}
void down(int ids, int id){
if(lazy[ids][id] != -1){
apply(ids, id << 1, lazy[ids][id]);
apply(ids, id << 1 | 1, lazy[ids][id]);
lazy[ids][id] = -1;
}
}
void update(int id, int l, int r, int u, int v, char c){
if(u <= l && r <= v){
FOR(ids, 0, num) apply(ids, id, (c == 'J' ? 0 : (c == 'O' ? 1 : 2)));
} else{
int mid = l + r >> 1;
FOR(ids, 0, num) down(ids, id);
if(u <= mid) update(id << 1, l, mid, u, v, c);
if(mid < v) update(id << 1 | 1, mid + 1, r, u, v, c);
FOR(ids, 0, num) is_equal[ids][id] = is_equal[ids][id << 1] & is_equal[ids][id << 1 | 1];
}
}
int main(){
setIO();
cin >> N >> S[0] >> S[1] >> S[2];
cin >> Q >> T;
candidates.pb(S[0]); //0
candidates.pb(S[1]); //1
candidates.pb(S[2]); //2
candidates.pb(f(S[0], S[1])); //3
candidates.pb(f(S[0], S[2])); //4
candidates.pb(f(S[1], S[2])); //5
candidates.pb(f(S[0], candidates[5]));
candidates.pb(f(S[1], candidates[4]));
candidates.pb(f(S[2], candidates[3]));
sort(all(candidates)); compact(candidates);
num = sz(candidates);
build(1, 0, N-1);
memset(lazy, -1, sizeof(lazy));
cout << (ok() ? "Yes\n" : "No\n");
while(Q--){
int l, r; char c;
cin >> l >> r >> c;
--l, --r;
update(1, 0, N-1, l, r, c);
cout << (ok() ? "Yes\n" : "No\n");
}
return 0;
}
