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 MX = 2e6 + 10, MD = 1e9 + 7;
int n, m, ans, tmp, u[MX], v[MX], edge, vertex, mx, mx2;
vector<int> adj[MX], V;
bool mark[MX];
void dfs(int v, int p = -1) {
  mark[v] = 1;
  vertex++;
  if (mx == -1)
    mx = v;
  else if (mx2 == -1)
    mx2 = v;
  else if (v > mx) {
    mx2 = mx;
    mx = v;
  } else if (v > mx2)
    mx2 = v;
  for (auto u : adj[v]) {
    edge++;
    if (mark[u]) continue;
    dfs(u, v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d%d", &v[i], &u[i]);
    V.push_back(u[i]);
    V.push_back(v[i]);
  }
  sort(V.begin(), V.end());
  V.resize(unique(V.begin(), V.end()) - V.begin());
  for (int i = 0; i < n; i++) {
    u[i] = lower_bound(V.begin(), V.end(), u[i]) - V.begin();
    v[i] = lower_bound(V.begin(), V.end(), v[i]) - V.begin();
    adj[u[i]].push_back(v[i]);
    adj[v[i]].push_back(u[i]);
  }
  for (int i = 0; i < n; i++) {
    if (mark[u[i]] == 0) {
      vertex = edge = 0;
      mx = mx2 = -1;
      dfs(u[i]);
      edge /= 2;
      if (vertex == edge)
        ans = max(ans, mx);
      else if (vertex - 1 == edge)
        ans = max(ans, mx2);
      else
        return cout << -1, 0;
    }
  }
  printf("%d", V[ans]);
}
