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 N = 1e5 + 5;
const short X = 4;
int n;
map<long long, int> cmp;
long long rev[2 * N];
short tp[N];
pair<long long, long long> q[N];
int cnt[4 * N];
short lazy[4 * N];
void down(int node, int l, int r) {
  if (lazy[node] == -1) {
    cnt[node << 1] = (l + r) / 2 - l + 1 - cnt[node << 1];
    cnt[node << 1 | 1] = r - (l + r) / 2 - cnt[node << 1 | 1];
    if (lazy[node << 1] == -1)
      lazy[node << 1] = X;
    else if (lazy[node << 1] <= 1)
      lazy[node << 1] ^= 1;
    else
      lazy[node << 1] = -1;
    if (lazy[node << 1 | 1] == -1)
      lazy[node << 1 | 1] = X;
    else if (lazy[node << 1 | 1] <= 1)
      lazy[node << 1 | 1] ^= 1;
    else
      lazy[node << 1 | 1] = -1;
  } else if (lazy[node] <= 1) {
    cnt[node << 1] = lazy[node] * ((l + r) / 2 - l + 1);
    cnt[node << 1 | 1] = lazy[node] * (r - (l + r) / 2);
    lazy[node << 1] = lazy[node << 1 | 1] = lazy[node];
  }
  lazy[node] = X;
  return;
}
void update(int node, int l, int r, int a, int b, short x) {
  if (a <= l && r <= b) {
    if (~x) {
      cnt[node] = x * (r - l + 1);
      lazy[node] = x;
    } else {
      cnt[node] = r - l + 1 - cnt[node];
      if (lazy[node] == -1)
        lazy[node] = X;
      else if (lazy[node] <= 1)
        lazy[node] ^= 1;
      else
        lazy[node] = -1;
    }
    return;
  }
  down(node, l, r);
  if (a <= (l + r) / 2) update(node << 1, l, (l + r) / 2, a, b, x);
  if (b > (l + r) / 2) update(node << 1 | 1, (l + r) / 2 + 1, r, a, b, x);
  cnt[node] = cnt[node << 1] + cnt[node << 1 | 1];
  return;
}
int query(int node, int l, int r) {
  if (l == r) return l;
  down(node, l, r);
  if (cnt[node << 1] < (l + r) / 2 - l + 1)
    return query(node << 1, l, (l + r) / 2);
  else
    return query(node << 1 | 1, (l + r) / 2 + 1, r);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> tp[i] >> q[i].first >> q[i].second;
    cmp[q[i].first] = cmp[q[i].second + 1] = 0;
  }
  int nums = 0;
  for (auto &v : cmp) {
    v.second = nums;
    rev[nums] = v.first;
    nums++;
  }
  if (rev[0] > 1) {
    while (n--) cout << 1 << '\n';
    return 0;
  }
  for (int i = 0; i < n; i++) {
    q[i].first = cmp[q[i].first];
    q[i].second = cmp[q[i].second + 1] - 1;
  }
  fill(lazy, lazy + 4 * N, X);
  for (int i = 0; i < n; i++) {
    update(1, 0, nums, q[i].first, q[i].second, 2 - tp[i]);
    if (cnt[1] == 0)
      cout << 1 << '\n';
    else
      cout << rev[query(1, 0, nums)] << '\n';
  }
  return 0;
}