#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int N = 35, M = 105, mod = 998244353;

int n, m, K;

int pwv[M][N];
int C[N][N];
int f[M][N][N][N];
int v[M];

inline int popcount(int x) {
	int ret = 0;
	while (x) ret += x & 1, x >>= 1;
	return ret;
}

int main()
{
	cin >> n >> m >> K;
	for (int i = 0; i <= m; i++) cin >> v[i];
	C[0][0] = 1;
	for (int i =1 ; i <= n; i++) {
		C[i][0] = 1;
		for (int j = 1; j <= i; j++) {
			C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
		}
	}
	for (int i = 0; i <= m; i++) {
		pwv[i][0] = 1;
		for (int j = 1; j <= n; j++) {
			pwv[i][j] = 1ll * pwv[i][j - 1] * v[i] % mod;
		}
	}
	
	// 解释一下f[i][j][k][p]的含义
	// 在S中确定了前i-1位，现在考虑i位，或者说，是用了v[0]~v[i-1]，现在用v[i]
	// 使用了a数组中的前j个数
	// k表示前i-1位中有多少个1
	// 向i位进位p个1
	f[0][0][0][0] = 1;
	for (int i = 0; i <= m; i++) {
	for (int j = 0; j <= n; j++) {
	for (int k = 0; k <= K; k++) { // mention
	for (int p = 0; p <= n / 2; p++) { // mention
	for (int t = 0; t <= n -j; t++) {
		f[i + 1][j + t][k + (t + p & 1)][(t + p) / 2] = 
		(f[i + 1][j + t][k + (t + p & 1)][(t + p) / 2] + 
		 1ll * f[i][j][k][p] * pwv[i][t] %mod * C[n - j][t] % mod
		) % mod;
		
	}}}}}
	
	int ans = 0;
	// we must need a f[m + 1][n][k][p]
	for (int k = 0; k <= K; k++) {
		for (int p = 0; p <= n / 2; p++) {
			if (popcount(p) + k <= K)
				ans = (ans + f[m + 1][n][k][p]) % mod;
		}
	}
	cout << ans << "\n";
	return 0;
}