Polycarp studies in Berland State University. Soon he will have to take his exam. He has to pass exactly n exams.

For the each exam i there are known two days: a_i — day of the first opportunity to pass the exam, b_i — day of the second opportunity to pass the exam (a_i < b_i). Polycarp can pass at most one exam during each day. For each exam Polycarp chooses by himself which day he will pass this exam. He has to pass all the n exams.

Polycarp wants to pass all the exams as soon as possible. Print the minimum index of day by which Polycarp can pass all the n exams, or print -1 if he cannot pass all the exams at all.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^6) — the number of exams.

The next n lines contain two integers each: a_i and b_i (1 ≤ a_i < b_i ≤ 10^9), where a_i is the number of day of the first passing the i-th exam and b_i is the number of day of the second passing the i-th exam.

Output

If Polycarp cannot pass all the n exams, print -1. Otherwise print the minimum index of day by which Polycarp can do that.

Examples

Input

2
1 5
1 7


Output

5


Input

3
5 13
1 5
1 7


Output

7


Input

3
10 40
40 80
10 80


Output

80


Input

3
99 100
99 100
99 100


Output

-1
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 1000001;
const int oo = 1e9;
int n, a[MaxN], b[MaxN], ans;
map<int, int> root;
int get_root(int u) {
  if (root[u] == 0) return u;
  return root[u] = get_root(root[u]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i] >> b[i];
  }
  for (int i = 1; i <= n; i++) {
    int p = get_root(a[i]), q = get_root(b[i]);
    ans = max(ans, min(p, q));
    if (ans == oo) break;
    if (p == q)
      root[p] = oo;
    else {
      if (p < q)
        root[p] = q;
      else
        root[q] = p;
    }
  }
  if (ans == oo)
    cout << "-1";
  else
    cout << ans;
  return 0;
}
