Description：

Solution：

Code：

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cmath>

#include<cctype>

#include<cstring>

using namespace std;

inline int rd()

{

	register int res = 0,f = 1;register char c = getchar();

	while(!isdigit(c)){if(c == '-')f = -1;c = getchar();}

	while(isdigit(c))res = (res << 1) + (res << 3) + c - '0',c = getchar();

	return res * f;

}

int n,m;

#define MAXN 1010

#define MAXM 17

#define MAX 15

char c[MAXM];

int toint(char c)

{

	if(c == 'N')return 0;if(c == 'O')return 1;

	if(c == 'I')return 2;

}

int S[MAXM];

int g[1 << MAX][3][3];

int dp[2][MAXN];

int bit(int s,int i){return ((s >> (i - 1)) & 1);}

int f[2][1 << MAX][3];

int ans[MAXM];

#define MOD 1000000007

int main()

{

	scanf("%d%d",&n,&m);

	scanf("%s",c + 1);

	for(int i = 1;i <= m;++i)S[i] = toint(c[i]);

	for(int i = 0;i < (1 << m);++i)

	{

		for(int x = 0;x < 3;++x)

		{

			for(int l = 0;l < 3;++l)

			{

				for(int k = 1;k <= m;++k)dp[0][k] = dp[0][k - 1] + bit(i,k);

				for(int k = 1;k <= m;++k)dp[1][k] = max(max(dp[0][k],dp[1][k - 1]),dp[0][k - 1] + (x == S[k]));

				int s = 0,len;

				for(int k = 1;k <= m;++k)s |= ((dp[1][k] - dp[1][k - 1]) << (k - 1));

				if(x == l)len = l + 1;

				else if(x == 0)len = 1;

				else len = 0;

				if(len == 3)g[i][x][l] = -1;

				else g[i][x][l] = s * 3 + len;

			}

		}

	}

	memset(f,0,sizeof(f));

	int cur = 0;

	f[0][0][0] = 1;

	for(int i = 1;i <= n;++i)

	{

		cur ^= 1;

		memset(f[cur],0,sizeof(f[cur]));

		for(int s = 0;s < (1 << m);++s)

		{

			for(int x = 0;x < 3;++x)

			{

				for(int l = 0;l < 3;++l)

				{

					int t = g[s][x][l] / 3,len = g[s][x][l] % 3;

					if(g[s][x][l] != -1)f[cur][t][len] = (f[cur][t][len] + f[cur ^ 1][s][l]) % MOD;

				}

			}

		}

	}

	memset(ans,0,sizeof(ans));

	for(int s = 0;s < (1 << m);++s)

	{

		int sum = 0;

		for(int k = 1;k <= m;++k)sum += (s >> (k - 1)) & 1;

		for(int l = 0;l < 3;++l)ans[sum] = (ans[sum] + f[cur][s][l]) % MOD;

	}

	for(int i = 0;i <= m;++i)printf("%d\n",ans[i]);	

	return 0;

}