#include <bits/stdc++.h>
#define isN (ss[i]=='N')
#define isA (ss[i]=='A')
#define isC (ss[i]=='C')
#define EM (ss[i]=='?')
using namespace std;
const int MAXN = 45,maxn = 2505;
bitset <maxn> dp[MAXN][MAXN][MAXN];
char ans[MAXN];
//dp[i][j][k][l]在1~i中有j个N，(i+1)~n中有k个C，能不能达到l个
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	string ss;
	int n,m;
	cin >> n >> m >> ss;
	ss = ' ' + ss;
	for (int i = 0; i <= n; i++) dp[0][0][i][0] = 1;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 0; j <= i; j++)
		{
			for (int k = 0; k <= n - i; k++)
			{
				if (!isN && !isA && !isC) dp[i][j][k] |= dp[i - 1][j][k];
				if ((isN || EM) && j) dp[i][j][k] |= dp[i - 1][j - 1][k];
				if (isA || EM) dp[i][j][k] |= (dp[i - 1][j][k] << (j * k));
				if (isC || EM) dp[i][j][k] |= dp[i - 1][j][k + 1];
			}
		}
	}
	int res = -1;
	for (int i = 0; i <= n; i++) if (dp[n][i][0][m]) res = i;
	if (res == -1)
	{
		cout << "-1\n";
		return 0;
	}
	for (int i = n,j = res,k = 0,l = m; i >= 1; i--)
	{
		if (isN || (EM && j && dp[i - 1][j - 1][k][l])) ans[i] = 'N',j--;
		else if (isA || (EM && j * k <= l && dp[i - 1][j][k][l - j * k])) ans[i] = 'A',l -= j * k;
		else if (isC || (EM && dp[i - 1][j][k + 1][l])) ans[i] = 'C',k++;
		else ans[i] = ss[i];
	}
	for (int i = 1; i <= n; i++) cout << ans[i];
	return 0;
}