CodeChef submission 136784 (C++ 4.3.2) plaintext list. Status: TLE, problem SNCK03, contest SNACKDWN. By SlodziakiUWr (SlodziakiUWr), 2009-11-21 23:57:05.
#include <cstdio>
#include <vector>
typedef long long LL;
#define MOD 1000000007
#define FORE(i,a,b) for(int i=(a);i<=(b);i++) 
 
int mod[3];
void EE (int a,int b,int c,int d,int e,int f){
 
	if(b==0){
 
   	  mod[0]=a;
 
      mod[1]=e;
 
      mod[2]=f;
 
   }
 
   else EE(b,a%b,e-(a/b)*c,f-(a/b)*d,c,d);
 
}
 
LL R[20005];
int odwrotnosc(int n){
	EE(n,MOD,0,1,1,0);
	return mod[1]%MOD;
 
}
 
 
inline int npok(int n, int k){
	return (R[n]*(LL)odwrotnosc((R[k]*R[n-k])%MOD))%MOD;
	
}
inline int modulo (int a){
    if (a >= MOD) a-= MOD;
    return a;
}
 
 
int n, k;
int A[30];
int DP[23][32][21000];
 
int gogo (int a, int b, int c)
{
    if (a + 1 == b) return 1;
    else
    {
        if (DP[a][b][c] == -1)
        {
            DP[a][b][c] = 0;
 
            if (a % 2 == b % 2)
            {
                // przypadki ze tylko dzielimy
                long long ret = 0;
                for (int h = a + 1; h < b; h += 2)
                {
                    ret += (npok ((A[b] - A[a] + 1 - c - 1 - (b!=k-1) - (a!=0)),(A[h] - A[a] + 1 - 1 - c - (a!=0)))) * (((long long)(gogo (a,h,c)) * gogo (h,b,0)) % MOD);
        			if (ret > 1000000000000000000LL) ret %= MOD;
                }
                DP[a][b][c] = ret % MOD;
            }
            else
            {
                // przypadki ze dzielimy i odcinamy przycinamy prefiks
                long long ret = 0;
 
                // dzielimy
                for (int h = a + 2; h < b; h += 2)
                {
   
                    ret +=  (npok ((A[b] - A[a] + 1 - c - 1 - (b!=k-1) - (a!=0)),(A[h] - A[a] + 1 - 1 - c - (a!=0)))) * (((long long)(gogo (a,h,c)) * gogo (h,b,0)) % MOD);
        			if (ret > 1000000000000000000LL) ret %= MOD;
                }
                // przycinamy prefiks
                if (A[a] + c + (a != 0) < A[a + 1])
                    ret += gogo (a,b,c + 1);
 
                DP[a][b][c] = ret % MOD;
            }
        }
 
        return DP[a][b][c];
    }
}
 
void doit ()
{
 
    scanf ("%d %d", &n, &k);
    for (int i = 0; i < k; ++i)
    {
        scanf ("%d", &A[i]);
        A[i]--;
    }
 
    for (int i = 0; i < k; ++i)
        for (int j = i + 1; j < k; ++j)
            for (int h = 0; h < A[j] - A[i] + 1; ++h)
                DP[i][j][h] = -1;
 
    printf ("%d\n", gogo (0, k - 1, 0));
}
 
int main ()
{
	R[0] = 1;
	FORE(i,1,20000){
		int j = i;
		R[i] = (R[i-1]*j)%MOD;
	}
 
 
    int tests;
    scanf ("%d", &tests);
 
    while (tests--) doit ();
 
    return 0;
}