def preprocess():
    '''1..9 の硬貨を使って、1..N を表す方法が何通りあるかを返す。
    dp[k][n]: 1..k円の硬貨を使って、n 円を表すパターン数 とすると、
    dp[k][0] = 1
    dp[k][j] = dp[j][j] (if k > j)
    dp[1][n] = 1
    dp[2][n] = sum(dp[1][n-2*i] for i in range(n//2))
             = dp[2][n-2] + dp[1][n]
             = dp[2][n-2] + 1
    dp[3][n] = sum(dp[2][n-3*i] for i in range(n//3))
             = dp[3][n-3] + dp[2][n]
    dp[k][n] = dp[k][n-k] + dp[k-1][n]
    '''
    M = 10**10
    N = M // 111111
    mod = 10**9 + 9
    dp = [1] * (N + 1)
    for k in range(2, 10):
        new_dp = [0] * (N + 1)
        for i in range(k):
            new_dp[i] = dp[i]
        for i in range(k, N + 1):
            new_dp[i] = (new_dp[i - k] + dp[i]) % mod
        dp = new_dp
    cumdp = []
    tmp = 0
    for dpi in dp:
        tmp += dpi
        tmp %= mod
        cumdp.append(tmp)
    return cumdp


def solve(m, dp):
    n = m // 111111
    return dp[n]


if __name__ == '__main__':
    dp = preprocess()
    T = int(input())
    for t in range(T):
        m = int(input())
        print(solve(m, dp))