//反向62
#include <iostream>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std; typedef long long ll;
// const int maxn = 1e5+5;
ll n;
ll a[];
ll dp[][]; //下标,前面是否4,是否有前导0, 是否有限制
ll dfs(ll pos, ll sta, ll pre, ll limit){
if(pos == -) return ;
if(!limit && dp[pos][sta] != -)
return dp[pos][sta];
int up = limit?a[pos]:;
ll ans = ;
for(int i = ;i <= up;i++){
if(pre == && i == ){
continue;
}
ans += dfs(pos-, i == , i, limit && i == a[pos]);
}
if(!limit) dp[pos][sta] = ans;
return ans;
} ll solve(ll x){
ll pos = ;
while(x){
a[pos++] = x%;
x /= ;
}
return dfs(pos-, , -, true);
} int main(){
memset(dp, -, sizeof dp);
int t;
scanf("%d", &t);
while(t--){
scanf("%lld", &n);
ll ans = solve(n);
printf("%lld\n", n - ans + );
}
return ;
}

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