#include <iostream>
#include <numeric>
#include <string>
#include <vector>
using namespace std;

#include <atcoder/modint>
using mint = atcoder::modint998244353;

vector<vector<mint>> ncr;
vector<mint> pow2;

mint solve(int N, int K, string X) {
    vector<mint> dp(N + 1);
    dp[N] = 1; // dp[# of candidate elements]

    for (auto c : X) {
        vector<mint> dpnxt(dp.size());
        const bool x = (c == '1');

        for (int i = K; i <= N; ++i) {
            for (int j = 0; j <= i; ++j) {
                if (x) {
                    // j : # of 1
                    if (i - j < K) dpnxt[i] += dp[i] * pow2[N - i] * ncr[i][j];
                }
                if (!x) {
                    // j : # of 0
                    if (j >= K) dpnxt[j] += dp[i] * pow2[N - i] * ncr[i][j];
                }
            }
        }
        dp = dpnxt;
    }

    return accumulate(dp.begin(), dp.end(), mint(0));
}

int main() {
    int N, M, K;
    string X;
    cin >> N >> M >> K >> X;

    ncr.assign(201, vector<mint>(201));
    for (int i = 0; i < int(ncr.size()); ++i) ncr[i][0] = 1;
    for (int i = 1; i < int(ncr.size()); ++i) {
        for (int j = 1; j < int(ncr[i].size()); ++j) ncr[i][j] = ncr[i - 1][j - 1] + ncr[i - 1][j];
    }
    pow2.assign(201, 1);
    for (int i = 1; i < int(pow2.size()); ++i) pow2[i] = pow2[i - 1] * 2;

    cout << solve(N, K, X).val() << '\n';
}