#include #include #include #include #include #include #include #include #include using namespace std; #define FOR(i, l, r) for(int i = l; i <= r; i++) #define REP(i, r, l) for(int i = r; i >= l; i--) typedef long long ll; double eps = 1e-6; const int inf = 0x3f3f3f3f; const int mod = 1e9+7; ll dp[2010][5];//[1]只有后一位,[2]后两位相同,[3]后三位相同 int main() { int T; scanf("%d",&T); memset(dp,0,sizeof(dp)); dp[1][1] = 26; dp[2][1] = 26*25, dp[2][2] = 26 ; FOR(i,3,2000) { dp[i][1] =1LL*25* ((dp[i-1][1] + dp[i-1][2] + dp[i-1][3])% mod); dp[i][1] %= mod; dp[i][2] = dp[i-1][1] %mod; dp[i][3] = dp[i-1][2]%mod; } while(T--) { int n; scanf("%d",&n); int ans = ((dp[n][1]+dp[n][2])%mod+dp[n][3])% mod; printf("%d\n",ans); } return 0; }