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 = 1e5 + 10;
long long xx[maxn << 2], x[maxn << 2];
int n, cnt, flag, c[maxn << 4], ans;
struct node {
  int t;
  long long l, r;
} a[maxn];
inline int pos(long long k) { return lower_bound(x + 1, x + cnt + 1, k) - x; }
void update(int h, int l, int r, int L, int R, int ty) {
  if (L <= l && r <= R && (ty != 3 || ty == 3 && c[h] >= 0)) {
    if (ty == 1) c[h] = 1;
    if (ty == 2) c[h] = 0;
    if (ty == 3 && c[h] >= 0) c[h] ^= 1;
    return;
  }
  if (c[h] >= 0) {
    c[h << 1] = c[h << 1 | 1] = c[h];
    c[h] = -1;
  }
  int mid = (l + r) >> 1;
  if (L <= mid) update(h << 1, l, mid, L, R, ty);
  if (mid + 1 <= R) update(h << 1 | 1, mid + 1, r, L, R, ty);
  if (c[h << 1] == c[h << 1 | 1] && c[h << 1] >= 0) c[h] = c[h << 1];
}
void query(int h, int l, int r) {
  if (c[h] == 1) return;
  if (c[h] == 0) {
    ans = l;
    flag = 1;
    return;
  }
  if (c[h] == -1) {
    int mid = (l + r) >> 1;
    query(h << 1, l, mid);
    if (flag) return;
    query(h << 1 | 1, mid + 1, r);
  }
}
int main() {
  cin >> n;
  for (register int i = 1; i <= n; i++) {
    cin >> a[i].t >> a[i].l >> a[i].r;
    xx[i * 4 - 3] = a[i].l - 1;
    xx[i * 4 - 2] = a[i].l;
    xx[i * 4 - 1] = a[i].r;
    xx[i * 4] = a[i].r + 1;
  }
  sort(xx + 1, xx + n * 4 + 1);
  for (register int i = 1; i <= n * 4; i++)
    if (xx[i] != xx[i - 1]) x[++cnt] = xx[i];
  x[cnt + 1] = x[cnt] + 1;
  cnt++;
  if (x[pos(1)] != 1) {
    x[++cnt] = 1;
    sort(x + 1, x + cnt + 1);
  }
  for (register int i = 1; i <= n; i++) {
    update(1, 1, cnt, pos(a[i].l), pos(a[i].r), a[i].t);
    flag = 0;
    ans = 0;
    query(1, 1, cnt);
    cout << x[ans] << endl;
  }
  return 0;
}