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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace ::std;
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a + b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long mul(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a * b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long pow_mod(long long a, long long b, long long m = MOD) {
  long long res = 1LL;
  a = a % m;
  while (b) {
    if (b & 1) res = mul(res, a, m);
    b >>= 1;
    a = mul(a, a, m);
  }
  return res;
}
long long fastexp(long long a, long long b) {
  long long res = 1LL;
  while (b) {
    if (b & 1) res = res * a;
    b >>= 1;
    a *= a;
  }
  return res;
}
int gcdExtendido(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtendido(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverso(int a, int m) {
  int x, y;
  int g = gcdExtendido(a, m, &x, &y);
  if (g != 1)
    return -1;
  else
    return (x % m + m) % m;
}
const int N = 100000 + 5;
int n;
long long l, r;
int ans[N];
long long f(int x) { return 2LL * n * x - 1LL * x * (x + 1); }
int main() {
  int t;
  scanf("%d", &(t));
  while (t--) {
    scanf("%d", &(n));
    scanf("%lld %lld", &(l), &(r));
    int len = r - l + 1;
    if (r == 1LL * n * (n - 1) + 1) {
      ans[r - l] = 1;
      r -= 1;
    }
    int lo = 1, hi = n;
    while (lo < hi) {
      int mi = lo + (hi - lo) / 2;
      if (f(mi) < l)
        lo = mi + 1;
      else
        hi = mi;
    }
    l -= f(lo - 1);
    r -= f(lo - 1);
    int pos = 0;
    int x = lo;
    int y = lo + (l + 1) / 2;
    bool goX = l & 1;
    for (int i = l; i <= r; i++) {
      if (goX) {
        ans[pos++] = x;
      } else {
        ans[pos++] = y;
        y += 1;
      }
      if (y == n + 1) {
        x += 1;
        y = x + 1;
      }
      goX = !goX;
    }
    for (int i = 0; i < len; i++) {
      printf("%d%c", ans[i], " \n"[i + 1 == len]);
    }
  }
  return 0;
}