1 //反向62
2 #include <iostream>
3 #include <algorithm>
4 #include <string>
5 #include <cstring>
6 #include <cstdio>
7 using namespace std;
8
9 typedef long long ll;
10 // const int maxn = 1e5+5;
11 ll n;
12 ll a[100];
13 ll dp[100][100];
14
15 //下标，前面是否4，是否有前导0， 是否有限制
16 ll dfs(ll pos, ll sta, ll pre, ll limit){
17 if(pos == -1) return 1;
18 if(!limit && dp[pos][sta] != -1)
19 return dp[pos][sta];
20 int up = limit?a[pos]:9;
21 ll ans = 0;
22 for(int i = 0;i <= up;i++){
23 if(pre == 4 && i == 9){
24 continue;
25 }
26 ans += dfs(pos-1, i == 4, i, limit && i == a[pos]);
27 }
28 if(!limit) dp[pos][sta] = ans;
29 return ans;
30 }
31
32 ll solve(ll x){
33 ll pos = 0;
34 while(x){
35 a[pos++] = x%10;
36 x /= 10;
37 }
38 return dfs(pos-1, 0, -1, true);
39 }
40
41 int main(){
42 memset(dp, -1, sizeof dp);
43 int t;
44 scanf("%d", &t);
45 while(t--){
46 scanf("%lld", &n);
47 ll ans = solve(n);
48 printf("%lld\n", n - ans + 1);
49 }
50 return 0;
51 }
