Problem: You are given a positive number x. Find the smallest positive integer number that has the sum of digits equal to x and all digits are distinct (unique).

Input

The first line contains a single positive integer t (1 ≤ t ≤ 50) — the number of test cases in the test. Then t test cases follow.

Each test case consists of a single integer number x (1 ≤ x ≤ 50).

Output

Output t answers to the test cases:

  * if a positive integer number with the sum of digits equal to x and all digits are different exists, print the smallest such number; 
  * otherwise print -1. 

Example

Input


4
1
5
15
50


Output


1
5
69
-1

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;
 
#define  GODSPEED       ios:: sync_with_stdio(0);cin.tie(0);cout.tie(0);cout<<fixed;cout<<setprecision(15);
#define  f              first               
#define  s              second
#define  newl           cout<<"\n";
#define  pb             push_back  
#define  mset(a,x)      memset(a,x,sizeof(a))
#define  debv(a)        for(auto it: a)cout<<it<<" ";newl;
#define  deb1(a)        cout<<a<<"\n";
#define  deb2(a,b)      cout<<a<<" "<<b<<"\n";
#define  deb3(a,b,c)    cout<<a<<" "<<b<<" "<<c<<"\n";
#define  deb4(a,b,c,d)  cout<<a<<" "<<b<<" "<<c<<" "<<d<<"\n";
#define  uniq(a)        a.resize(unique(a.begin(), a.end()) - a.begin());
#define  all(a)         a.begin(),a.end()           
 
typedef  long long             ll;
typedef  long double           ld;
typedef  pair<ll,ll>           pll;
typedef  vector<ll>            vll;
typedef  vector<pll>           vpll;

const ll   N     =  3e5+5;
const ll   mod   =  998244353;
const ll   INF   =  0x7f7f7f7f7f7f7f7f;
const int  INFi  =  0x7f7f7f7f;
const ll   LEVEL =  log2(N)+1;

int n, a[N];
set <int> s;

void rec(int sum, int no){
	if(a[sum] == -1) a[sum] = no;
	a[sum] = min(a[sum], no);
	for(int i = 1; i <= 9; i++){
		if(s.find(i) == s.end()) continue;
		s.erase(i);
		rec(sum + i, no * 10 + i);
		s.insert(i);
	}
}

void solve(){
	cin >> n;
	deb1(a[n])
}

int main(){
	GODSPEED;
	mset(a, -1);
	for(int i = 1; i <= 9; s.insert(i++));
	rec(0, 0);
	int test = 1;
	cin >> test;
	for(int i = 1; i <= test; i++){
		solve();
	}
	#ifndef ONLINE_JUDGE
	cout<<"\nTime Elapsed: " << 1.0*clock() / CLOCKS_PER_SEC << " sec\n";
	#endif
}