I didn’t solve this problem in the club.

There’s a straightforward dynamic progamming where $f[i][j]$ represents the minimum number of cuts required for $s[0..i]$ and $t[0..j]$ but this would obviously exceed the time limit.
A direct idea is swap the state $j$ and the result:
$\text{dp}[\text{pref}][\text{cnt}]$ represents the length of the longest prefix of $t$ that can be formed by dividing $s[0..\text{pref}]$ into $\text{cnt}$ parts. However this doesn’t solve the time issue, as the number of states is still $O(n^2)$.

Actually, the intermediate states are completely useless. What’s ultimately required is whether the complete $s$ can be cut into $t$, so some states can be skipped. When there is a common substring, it’s not necessary to update every substring length state, only the final state (the longest common substring length) matter. State transition equation:

$$dp[\text{pref}+1][\text{cnt}]=\max\{dp[\text{pref}][\text{cnt}]\}$$ $$dp[\text{pref}+\text{len}][\text{cnt}+1]=\max\{dp[\text{pref}][\text{cnt}]+\text{len}\}$$

Where len represents the length of the longest common substring starting from $\text{pref} + 1$, and use string double hashing and binary search, total time $O(nx\log)$.

My initial idea was to add an extra dimension to dp, indicating whether the last cut was made at the far right, but this is actually a redundant state, regardless of where the last cut was made, the current state would definitely be updated by some previous states.

// compile: make data
// run: ./data < data.in
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#ifdef LOCAL
#include <debug/codeforces.h>
#define debug(x...) _debug_print(#x, x);
#define Debug(x...) _debug_print_format(#x, x);
std::ifstream terminal("/dev/tty");
#define PP cerr<<"\033[1;30mpause...\e[0m",terminal.ignore();
#else
#define debug(x...)
#define Debug(x...)
#define PP
#endif
template<typename...Args> void print_(Args...args){((cout<<args<<" "),...)<<endl;}
#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)
#define sz(v) ((int)(v).size())
#define print(...) print_(__VA_ARGS__);
#define FIND(a, x) ((find(a.begin(),a.end(),(x))!=a.end())?1:0)
#define cmin(x,...) x=min({(x),__VA_ARGS__})
#define cmax(x,...) x=max({(x),__VA_ARGS__})
#define INTMAX (int)(9223372036854775807)
#define INF (int)(1152921504606846976)
#define NaN (int)(0x8b88e1d0595d51d1)
#define double long double
#define int long long
#define uint unsigned long long
#define MAXN 200010

template <char F, int B> struct strdhash {
    const pair<int, int> M = {999999893, 999999739};
    string s;
    int len;
    vector<pair<int, int>> pw, hs;
    strdhash(string S): s(S), len(S.size()) {
        pw.assign(len, {0, 0});
        pw[0] = {1, 1};
        rep(i, 1, len) pw[i].first = pw[i-1].first * B % M.first, pw[i].second = pw[i-1].second * B % M.second;
        hs.assign(len+1, {0, 0});
        rep(i, 0, len) hs[i+1].first = (hs[i].first * B + s[i] - F) % M.first, hs[i+1].second = (hs[i].second * B + s[i] - F) % M.second;
    }
    pair<int, int> hash(int l, int r) {
        if (!l) return hs[r+1];
        pair<int, int> res;
        res.first = (hs[r+1].first - hs[l].first * pw[r-l+1].first % M.first + M.first) % M.first;
        res.second = (hs[r+1].second - hs[l].second * pw[r-l+1].second % M.second + M.second) % M.second;
        return res;
    }
};

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

    int n, m, x;
    string s, t;
    cin >> n >> s >> m >> t >> x;
    s = "#" + s, t = "@" + t;
    strdhash hs = strdhash<'a', 26>(s);
    strdhash ht = strdhash<'a', 26>(t);
    vector<vector<int>> dp(n+1, vector<int>(x+2, 0));
    auto strc = [&](int S, int T, int len) {
        return hs.hash(S, S+len-1) == ht.hash(T, T+len-1);
    };
    auto bs = [&](int S, int T) {
        int l = 0, r = min(n-S+1, m-T+1);
        int ans = -1;
        while (l <= r) {
            int mid = (l + r) / 2;
            if (strc(S, T, mid)) ans = mid, l = mid + 1;
            else r = mid - 1;
        }
        return ans;
    };
    rep(i, 0, n) {
        rep(j, 0, x+1) {
            cmax(dp[i+1][j], dp[i][j]);
            int len = bs(i+1, dp[i][j]+1);
            if (~len) cmax(dp[i+len][j+1], dp[i][j] + len);
        }
    }
    cout << (dp[n][x] == m ? "YES" : "NO") << endl;

    return 0;
}