Problem: There are n segments [l_i, r_i] for 1 ≤ i ≤ n. You should divide all segments into two non-empty groups in such way that there is no pair of segments from different groups which have at least one common point, or say that it's impossible to do it. Each segment should belong to exactly one group.

To optimize testing process you will be given multitest.

Input

The first line contains one integer T (1 ≤ T ≤ 50000) — the number of queries. Each query contains description of the set of segments. Queries are independent.

First line of each query contains single integer n (2 ≤ n ≤ 10^5) — number of segments. It is guaranteed that ∑{n} over all queries does not exceed 10^5.

The next n lines contains two integers l_i, r_i per line (1 ≤ l_i ≤ r_i ≤ 2 ⋅ 10^5) — the i-th segment.

Output

For each query print n integers t_1, t_2, ..., t_n (t_i ∈ \{1, 2\}) — for each segment (in the same order as in the input) t_i equals 1 if the i-th segment will belongs to the first group and 2 otherwise.

If there are multiple answers, you can print any of them. If there is no answer, print -1.

Example

Input


3
2
5 5
2 3
3
3 5
2 3
2 3
3
3 3
4 4
5 5


Output


2 1 
-1
1 1 2 

Note

In the first query the first and the second segments should be in different groups, but exact numbers don't matter.

In the second query the third segment intersects with the first and the second segments, so they should be in the same group, but then the other group becomes empty, so answer is -1.

In the third query we can distribute segments in any way that makes groups non-empty, so any answer of 6 possible is correct.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
const long long N = 200005;
vector<long long> vis(N);
vector<vector<long long>> adj(N);
void dfs(long long v) {
  vis[v] = 1;
  for (auto i : adj[v]) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  return;
}
bool isPrime(long long n) {
  if (n < 2) return false;
  for (long long i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res % mod;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
long long max(long long a, long long b) {
  long long ans = a > b ? a : b;
  return ans;
}
long long min(long long a, long long b) {
  long long ans = a < b ? a : b;
  return ans;
}
long long root(long long a[], long long i) {
  while (a[i] != i) {
    a[i] = a[a[i]];
    i = a[i];
  }
  return i;
}
void unionn(long long a[], long long i, long long j) {
  long long root_i = root(a, i);
  long long root_j = root(a, j);
  a[root_i] = root_j;
  return;
}
void fun() {}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  fun();
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    long long c1 = 0, c2 = 0;
    long long n;
    cin >> n;
    map<long long, pair<long long, long long>> m;
    map<pair<long long, long long>, long long> m1;
    long long maxx = 0;
    vector<pair<long long, long long>> a(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i].first >> a[i].second;
      m[i] = a[i];
    }
    sort(a.begin(), a.end());
    for (long long i = 0; i < n; i++) {
      if (!c1) {
        m1[a[i]] = 1;
        c1 = 1;
        maxx = max(maxx, a[i].second);
      } else {
        if (a[i].first > maxx) {
          if (m1[a[i - 1]] == 1) {
            m1[a[i]] = 2;
            c2 = 1;
          } else {
            m1[a[i]] = 1;
          }
          maxx = max(maxx, a[i].second);
        } else {
          m1[a[i]] = m1[a[i - 1]];
          maxx = max(maxx, a[i].second);
        }
      }
    }
    if (!c2) {
      cout << "-1"
           << "\n";
      continue;
    }
    for (long long i = 0; i < n; i++) cout << m1[m[i]] << " ";
    cout << "\n";
  }
  return 0;
}