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:
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#include <complex>
#include <queue>
#include <set>
#include <unordered_set>
#include <list>
#include <chrono>
#include <random>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <unordered_map>
#include <stack>
#include <iomanip>
#include <fstream>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<double, double> pdd;
typedef vector<ll> vll;
typedef vector<int> vi;
ll MOD = 998244353;
double eps = 1e-12;
#define forn(i, e) for (ll i = 0; i < e; i++)
#define forsn(i, s, e) for (ll i = s; i < e; i++)
#define rforn(i, s) for (ll i = s; i >= 0; i--)
#define rforsn(i, s, e) for (ll i = s; i >= e; i--)
#define ln "\n"
#define dbg(x) cout << #x << " = " << x << ln
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define INF 2e18
#define fast_cin()                    \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())

void solve()
{
    int n;
    cin >> n;
    if (n > 45)
    {
        cout << -1 << endl;
        return;
    }
    if (n < 10)
    {
        cout << n << endl;
        return;
    }

    string ans = "";

    for (int i = 9; i >= 1; i--)
    {
        if (n <= 9 && n <= i)
        {
            ans += to_string(n);
            n = 0;
            break;
        }
        else
        {
            ans += to_string(i);
            n -= i;
        }
    }

    if (n)
    {
        cout << -1 << endl;
        return;
    }
    reverse(all(ans));

    cout << ans << endl;
}
int main()
{
    fast_cin();
    ll t = 1;
    cin >> t;
    for (int it = 1; it <= t; it++)
    {
        // cout << "Case #" << it + 1 << ": ";
        solve();
    }
    return 0;
}