#include<cstdio>#include<cstring>#include<cstdlib>#include<cctype>#include<cmath>#include<iostream>#include<sstream>#include<iterator>#include<algorithm>#include<string>#include<vector>#include<set>#include<map>#include<stack>#include<deque>#include<queue>#include<list>#define lowbit(x) (x & (-x))const double eps = 1e-8;inline int dcmp(double a, double b){    if(fabs(a - b) < eps) return 0;    return a > b ? 1 : -1;}typedef long long LL;typedef unsigned long long ULL;const int INT_INF = 0x3f3f3f3f;const int INT_M_INF = 0x7f7f7f7f;const LL LL_INF = 0x3f3f3f3f3f3f3f3f;const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};const int MOD = 1e9 + 7;const double pi = acos(-1.0);const int MAXN = 10000 + 10;const int MAXT = 10000 + 10;using namespace std;int digit[30];LL dp[30][2];LL dfs(int len, bool state, bool limit){//len--当前位，从高到低枚举，state--上一位的状态，limit--当前位的数字是否有限制    if(!len) return 1;    if(!limit && dp[len][state] != -1) return dp[len][state];//!limit--记录重复枚举的数的个数，以便记忆化搜索    LL ans = 0, up = limit ? digit[len] : 9;    for(int i = 0; i <= up; ++i){        if(state && i == 9) continue;        ans += dfs(len - 1, i == 4, limit && i == up);    }    if(!limit) dp[len][state] = ans;//记录截止到当前位且当前位无限制时满足条件的数的个数    return ans;}LL solve(LL x){    int cnt = 0;    while(x){        digit[++cnt] = x % 10;        x /= 10;    }    return dfs(cnt, false, true);}int main(){    int T;    scanf("%d", &T);    memset(dp, -1, sizeof dp);    while(T--){        LL N;        scanf("%lld", &N);        printf("%lld\n", N - (solve(N) - solve(0)));    }    return 0;}