Help: What is the Time Complexity of this?

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I was attempting the E problem of round #878 Div 3. I am getting TLE in my code but it seems to me that the Time complexity of the code if O(T * (t + q)) which should get accepted. Please tell me the mistake in the code.

Approach: I am storing the position which are unblocked at time (relative to current time) in a queue. The string a and b are just a copy of s1 and s2 which will be blocked and unblocked throughout the process (s1 and s2 are not altered).

Problem: Div 3 E
My submission: Submission

Code

#include <bits/stdc++.h>
#define rep(i, n) for (long long i = 0; i < n; ++i)
#define coutyes cout << "YES" << nl
#define coutno cout << "NO" << nl
#define IOS                       \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0)
#define nl '\n'
using namespace std;
using ll = long long;
 
void solve()
{
    string a, b, s1, s2; cin >> s1 >> s2;
    a = s1, b = s2;
    ll t, q; cin >> t >> q;
    queue<ll> qu;
    rep(i,t) qu.push(-1);
    while(q--){
        ll x; cin >> x;

        if(qu.front()==-1) qu.pop();
        else{
            ll k=qu.front();
            a[k]=s1[k], b[k]=s2[k];
            qu.pop();
        }

        if(x==3){
            qu.push(-1);
            if(a==b)coutyes;
            else coutno;
        }
        else if(x==2){
            qu.push(-1);
            ll p, q, r, s; cin >> p >> q >> r >> s;
            if(p==1 && r==1)
                swap(a[q - 1], a[s - 1]), swap(s1[q - 1], s1[s - 1]);
            else if(p==1 && r==2)
                swap(a[q - 1], b[s - 1]), swap(s1[q - 1], s2[s - 1]);
            else if(p==2 && r==1)
                swap(b[q - 1], a[s - 1]), swap(s2[q - 1], s1[s - 1]);
            else
                swap(b[q - 1], b[s - 1]), swap(s2[q - 1], s2[s - 1]);
        }
        else{
            ll y; cin>>y;
            qu.push(y - 1); a[y-1]='A'; b[y-1]='A';
        }
    }       
}
signed main()
{
    IOS;
    int t=1; cin >> t;
    while(t--) solve();
}