#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef complex<double> P;
typedef pair<int,int> pii;
#define REP(i,n) for(ll i=0;i<n;++i)
#define REPR(i,n) for(ll i=1;i<n;++i)
#define FOR(i,a,b) for(ll i=a;i<b;++i)

#define DEBUG(x) cout<<#x<<": "<<x<<endl
#define DEBUG_VEC(v) cout<<#v<<":";REP(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

#define MOD (ll)(1e9+7)
#define ADD(a,b) a=((a)+(b))%MOD
#define FIX(a) ((a)%MOD+MOD)%MOD

// keta less bef2 bef1
ll dp[20][2][11][11];

ll count(ll bound){
  // keta dp
  string s = to_string(bound);
  int n = s.size();
  REP(i,n+1)REP(j,2)REP(k,11)REP(l,11)dp[i][j][k][l] = 0;
  dp[0][0][10][10] = 1;
  REP(keta,n){
    int d = s[keta]-'0';
    REP(less,2)REP(bef2,11)REP(bef1,11){
      REP(k,10){
        if(less || k<=d){
          int nl = less || k<d;
          if(bef1==10 || (bef2==10&&bef1!=k) || (bef2<bef1 && bef1>k && bef2!=k) || (bef2>bef1 && bef1<k && bef2!=k)){
            // ok
            dp[keta+1][nl][bef1][(bef1==10&&k==0)?10:k] += dp[keta][less][bef2][bef1];
          }
        }
      }
    }
  }
  ll ans = 0;
  REP(less,2)REP(bef2,10)REP(bef1,10){
    ans += dp[n][less][bef2][bef1];
  }
  return ans;
}

void solve(){
  ll k;
  scanf("%lld",&k);
  k += count(100);
  ll low = 0, high = 1e17;
  while(low+1<high){
    ll mid = (low+high)/2;
    ll cnt = count(mid);
    if(cnt>=k) high = mid;
    else low = mid;
  }
  printf("%lld\n",high);
}

int main(){
  int t;
  scanf("%d",&t);
  while(t--)solve();
  return 0;
}