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;
long long n, opt[400050], l[400050], r[400050];
long long a[400050], tot, q;
int t[400050 << 2], f[400050 << 2];
int get(long long x) { return lower_bound(a + 1, a + q + 1, x) - a; }
void pushup(int id) {
  if (t[id << 1] == t[id << 1 | 1])
    t[id] = t[id << 1];
  else
    t[id] = 2;
}
void down(int id) {
  if (f[id] == 0) return;
  if (f[id] == 3) {
    if (t[id << 1] != 2) t[id << 1] ^= 1;
    if (f[id << 1] != 3) {
      if (f[id << 1] == 0)
        f[id << 1] = 3;
      else
        f[id << 1] = (f[id << 1] == 1) ? 2 : 1;
    } else
      f[id << 1] = 0;
    if (t[id << 1 | 1] != 2) t[id << 1 | 1] ^= 1;
    if (f[id << 1 | 1] != 3) {
      if (f[id << 1 | 1] == 0)
        f[id << 1 | 1] = 3;
      else
        f[id << 1 | 1] = (f[id << 1 | 1] == 1) ? 2 : 1;
    } else
      f[id << 1 | 1] = 0;
  } else if (f[id] <= 2)
    t[id << 1] = t[id << 1 | 1] = (f[id] == 1 ? 1 : 0),
            f[id << 1] = f[id << 1 | 1] = f[id];
  f[id] = 0;
}
void update(int id, int l, int r, int x, int y, int d) {
  if (l > y || r < x) return;
  if (l >= x && r <= y) {
    if (d <= 2)
      t[id] = (d == 1) ? 1 : 0, f[id] = d;
    else {
      if (t[id] != 2) t[id] ^= 1;
      if (f[id] == 0)
        f[id] = 3;
      else if (f[id] <= 2)
        f[id] = (f[id] == 1) ? 2 : 1;
      else
        f[id] = 0;
    }
    return;
  }
  int mid = (l + r) >> 1;
  down(id);
  update(id << 1, l, mid, x, y, d);
  update(id << 1 | 1, mid + 1, r, x, y, d);
  pushup(id);
}
int query(int id, int l, int r) {
  if (l == r) return l;
  int mid = (l + r) >> 1;
  down(id);
  int ans = 0;
  if (t[id << 1] != 1)
    ans = query(id << 1, l, mid);
  else
    ans = query(id << 1 | 1, mid + 1, r);
  pushup(id);
  return ans;
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld%lld", &opt[i], &l[i], &r[i]);
    if (l[i] != 1) a[++tot] = l[i] - 1;
    a[++tot] = l[i], a[++tot] = r[i] + 1, a[++tot] = r[i];
  }
  a[++tot] = 1;
  sort(a + 1, a + tot + 1);
  q = unique(a + 1, a + tot + 1) - a - 1;
  for (int i = 1; i <= n; i++) {
    update(1, 1, q, get(l[i]), get(r[i]), opt[i]);
    printf("%lld\n", a[query(1, 1, q)]);
  }
  return 0;
}