#include <stdio.h>
#include <queue>

using namespace std;

// It's like a BFS algorithm, where you have 9 roots for 9 trees (1, 2, ..., 9).
// For each tree, you scan it level by level,
// and you generate the two children of every number,
// add them to the queue, and remove the current number at q.front().

// When you generate a children, make sure that the digit is >= 0 and <= 9.
// So, If currNum = 9, therefore, two possible children 9[8] (valid) and 9[10] (invalid).

// This algorithm will take O(10 * LogN), where Log is to the base 10. Because in every tree level we multiply by 10.

int main() {
	int t;
    scanf("%d",&t);
    while(t--) {
        long long n;
        scanf("%lld",&n);

        if(n <= 9) {
            printf("%d\n", -1); continue;
        }

        queue<long long> q;
        for(int i = 1; i <= 9; i++)
            q.push(i);

        while(true) {
            long long currNum = q.front();
            int lastDigit = currNum % 10;

            if(currNum > n)
                break;
            else if(currNum > 9 && currNum <= n)
                printf("%lld ", currNum);

            if(lastDigit - 1 >= 0){
                q.push(currNum * 10 + (lastDigit - 1));
            }
            if(lastDigit + 1 <= 9){
                q.push(currNum * 10 + (lastDigit + 1));
            }
            q.pop();
        }

        printf("\n");
    }
	return 0;
}