Problem: You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.

You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).

We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 — a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.

Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.

Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — the number of vertices in K_n, and segment of the cycle to print.

It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.

Output

For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.

Example

Input


3
2 1 3
3 3 6
99995 9998900031 9998900031


Output


1 2 1 
1 3 2 3 
1 

Note

In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.

In the third test case, it's quite obvious that the cycle should start and end in vertex 1.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.StringTokenizer;

import static java.lang.Double.parseDouble;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.System.exit;


public class Solution {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;
    static boolean isLocal = false;

    void Case() throws IOException {
        long n = nextLong();
        long l = nextLong(), r = nextLong();
        long sum = 0L;
        int i;
        for (i = 1; i <= n; i++) {
            sum += 2 * (n - i);
            if (l <= sum) {
                sum -= 2 * (n - i);
                sum++;
                break;
            }
        }
        for (long ii = i; ii <= n - 1; ii++) {
            for (long j = ii + 1; j <= n; j++) {
                if (sum % 2 == 1) {
                    if (sum >= l && sum <= r) out.print(ii + " ");
                    sum++;
                }
                if (sum % 2 == 0) {
                    if (sum >= l && sum <= r) out.print(j + " ");
                    sum++;
                }
                if (sum > r) {
                    out.println();
                    return;
                }
            }
        }
        out.println(1);
    }

    void solve() throws Exception {
        int t = nextInt();
        while (t-- > 0)
            Case();
    }

    int[] sort(int[] arr) {
        sort(arr, 0, arr.length - 1);
        return arr;
    }

    void sort(int arr[], int l, int r) {
        if (l < r) {
            int m = (l + r) / 2;
            sort(arr, l, m);
            sort(arr, m + 1, r);
            merge(arr, l, m, r);
        }
    }

    void merge(int arr[], int l, int m, int r) {
        int n1 = m - l + 1;
        int n2 = r - m;
        int L[] = new int[n1];
        int R[] = new int[n2];
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];
        int i = 0, j = 0;
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }


    private int[] na(int n) throws IOException {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) a[i] = nextInt();
        return a;
    }

    private long[] nal(int n) throws IOException {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) a[i] = nextLong();
        return a;
    }

    int nextInt() throws IOException {
        return parseInt(next());
    }

    long nextLong() throws IOException {
        return parseLong(next());
    }

    double nextDouble() throws IOException {
        return parseDouble(next());
    }

    String next() throws IOException {
        while (tok == null || !tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    public static void main(String[] args) throws Exception {
        try {
            if (isLocal) {
                in = new BufferedReader(new FileReader("src/tests/sol.in"));
                out = new PrintWriter(new BufferedWriter(new FileWriter("src/tests/sol.out")));
            } else {
                in = new BufferedReader(new InputStreamReader(System.in));
                out = new PrintWriter(new OutputStreamWriter(System.out));
            }

            //long lStartTime = System.currentTimeMillis();
            new Solution().solve();
            //long lEndTime = System.currentTimeMillis();
            //out.println("Elapsed time in seconds: " + (double)(lEndTime - lStartTime) / 1000.0);
            in.close();
            out.close();
        } catch (Throwable e) {
            e.printStackTrace();
            exit(1);
        }
    }
}