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// clang-format off
#include<bits/stdc++.h>
using namespace std;
using LL=long long;
#define FOR(i,l,r) for(auto i=(l);i<=(r);++i)
#define REP(i,n) FOR(i,0,(n)-1)
#define ssize(x) int(x.size())
template<class A,class B>auto&operator<<(ostream&o,pair<A,B>p){return o<<"("<<p.first<<", "<<p.second<<")";}
template<class T>auto operator<<(ostream&o,T x)->decltype(x.end(),o){o<<"{";int i=0;for(auto e:x)o<<(", ")+2*!i++<<e;return o<<"}";}
#ifdef DEBUG
#define debug(X...)cerr<<"["#X"]: ",[](auto...$){((cerr<<$<<"; "),...)<<"\n";}(X)
#else
#define debug(...) {}
#endif
// clang-format on

int
main()
{
    cin.tie(0)->sync_with_stdio(0);
    int N, K, T;
    string seq;
    cin >> N >> K >> T >> seq;

    // Initial dp values
    const short NINF = numeric_limits<short>::min();
    vector<vector<vector<short>>> dp(
      N + 1, vector<vector<short>>(2, vector<short>(K + 1, NINF)));
    dp[0][0][0] = 0;

    // Calculate dp
    int drive_penalty = 0;
    FOR (n, 1, N) {
        const auto& ch    = seq[n - 1];
        bool some_meeting = ch != '3';

        // Update penalty for finishing a drive at n
        drive_penalty += some_meeting;
        if (n - T >= 1) drive_penalty -= seq[n - T - 1] != '3';

        // Logging
        auto region = seq.substr(max(0, n - T), min(T, n));
        debug(n, region.c_str(), drive_penalty);

        // Maybe use car
        if (n - T >= 0) {
            FOR (k, drive_penalty, K) {
                dp[n][0][k] = dp[n - T][1][k - drive_penalty];
                dp[n][1][k] = dp[n - T][0][k - drive_penalty];
            }
        }

        switch (seq[n - 1]) {
            case '1':  // meeting at work
                // Politely work at work
                FOR (k, 0, K) dp[n][1][k] = max(dp[n][1][k], dp[n - 1][1][k]);

                // Ignore the meeting and do potyczki at home
                FOR (k, 1, K)
                    dp[n][0][k]
                      = max<short>(dp[n][0][k], dp[n - 1][0][k - 1] + 1);
                break;
            case '2':
                // Politely work at work or home
                FOR (k, 0, K) dp[n][0][k] = max(dp[n][0][k], dp[n - 1][0][k]);
                FOR (k, 0, K) dp[n][1][k] = max(dp[n][1][k], dp[n - 1][1][k]);

                // Ignore the meeting and do potyczki at home
                FOR (k, 1, K)
                    dp[n][0][k]
                      = max<short>(dp[n][0][k], dp[n - 1][0][k - 1] + 1);
                break;
            default:
                // Do potyczki at home in free time
                FOR (k, 0, K) dp[n][0][k] = max<short>(dp[n][0][k], dp[n - 1][0][k] + 1);

                // Politely work at work
                FOR (k, 0, K) dp[n][1][k] = max(dp[n][1][k], dp[n - 1][1][k]);
                break;
        }

        debug(n, dp[n]);
    }

    auto score = max<short>(-1, *max_element(dp.back()[0].begin(), dp.back()[0].end()));
    cout << score << "\n";
    return 0;
}