Problem: You are given a set of integer numbers, initially it is empty. You should perform n queries.

There are three different types of queries: 

  * 1 l r — Add all missing numbers from the interval [l, r]
  * 2 l r — Remove all present numbers from the interval [l, r]
  * 3 l r — Invert the interval [l, r] — add all missing and remove all present numbers from the interval [l, r]



After each query you should output MEX of the set — the smallest positive (MEX ≥ 1) integer number which is not presented in the set.

Input

The first line contains one integer number n (1 ≤ n ≤ 105).

Next n lines contain three integer numbers t, l, r (1 ≤ t ≤ 3, 1 ≤ l ≤ r ≤ 1018) — type of the query, left and right bounds.

Output

Print MEX of the set after each query.

Examples

Input

3
1 3 4
3 1 6
2 1 3


Output

1
3
1


Input

4
1 1 3
3 5 6
2 4 4
3 1 6


Output

4
4
4
1

Note

Here are contents of the set after each query in the first example:

  1. {3, 4} — the interval [3, 4] is added 
  2. {1, 2, 5, 6} — numbers {3, 4} from the interval [1, 6] got deleted and all the others are added 
  3. {5, 6} — numbers {1, 2} got deleted 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
const int MOD = 1000000007;
map<long long, bool> seq;
void upd(int t, long long l, long long r) {
  auto it = seq.lower_bound(l);
  int prev = -1;
  if (it != seq.begin() && (it == seq.end() || it->first != l)) {
    it--;
    prev = it->second;
    it++;
  }
  int last = -1;
  while (it != seq.end() && it->first <= r) {
    last = it->second;
    if (t == 3) {
      it->second = !it->second;
      it++;
    } else {
      it = seq.erase(it);
    }
  }
  if (t == 1) {
    seq[l] = 1;
  } else if (t == 2) {
    seq[l] = 0;
  }
  if (prev != -1) {
    if (t == 3) seq[l] = !prev;
    if (last == -1) last = prev;
  }
  if (last == -1) last = 0;
  if (it == seq.end() || it->first > r + 1) {
    seq[r + 1] = last;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  seq[1] = 0;
  while (n--) {
    long long t, l, r;
    cin >> t >> l >> r;
    upd(t, l, r);
    auto it = seq.begin();
    while (it != seq.end() && it->second == 1) it++;
    cout << it->first << endl;
  }
}