You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.

Segment [l1, r1] lies within segment [l2, r2] iff l1 ≥ l2 and r1 ≤ r2.

Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Input

The first line contains one integer n (1 ≤ n ≤ 3·105) — the number of segments.

Each of the next n lines contains two integers li and ri (1 ≤ li ≤ ri ≤ 109) — the i-th segment.

Output

Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.

Examples

Input

5
1 10
2 9
3 9
2 3
2 9


Output

2 1


Input

3
1 5
2 6
6 20


Output

-1 -1

Note

In the first example the following pairs are considered correct:

  * (2, 1), (3, 1), (4, 1), (5, 1) — not even touching borders; 
  * (3, 2), (4, 2), (3, 5), (4, 5) — touch one border; 
  * (5, 2), (2, 5) — match exactly. 
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m, l = 0, p = 0, i, flag = 0, k, j, r = 0, q = 0, t, curr = 0, d;
  cin >> n;
  vector<pair<long long, long long>> v1, v2;
  long long a[n], b[n];
  vector<pair<long long, pair<long long, long long>>> v;
  map<long long, vector<long long>> m3, m1, m6, m4;
  map<long long, long long> m2, m5;
  for (long long i = 0; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v.push_back({x, {y, i}});
    a[i] = x;
    b[i] = x;
    v2.push_back({y, i});
    m1[x].push_back(i + 1);
    m2[x]++;
    m5[y]++;
    m4[y].push_back(i + 1);
    m3[x].push_back(y);
    m6[y].push_back(x);
    if (m5[y] == 2 && flag == 0) {
      flag = 1;
      if (m6[y][0] <= m6[y][1]) {
        v1.push_back({m4[y][1], m4[y][0]});
      } else {
        v1.push_back({m4[y][0], m4[y][1]});
      }
    }
    if (m2[x] == 2 && flag == 0) {
      flag = 1;
      if (m3[x][0] >= m3[x][1]) {
        v1.push_back({m1[x][1], m1[x][0]});
      } else {
        v1.push_back({m1[x][0], m1[x][1]});
      }
    }
  }
  if (flag == 1) {
    cout << v1[0].first << " " << v1[0].second;
    return 0;
  }
  if (flag == 0) {
    sort((v2).begin(), (v2).end());
    sort(b, b + n);
    j = v2[n - 1].first;
    long long r1 = v2[n - 1].second;
    if (a[r1] != b[n - 1]) {
      cout << v2[n - 2].second + 1 << " " << v2[n - 1].second + 1;
      return 0;
    } else {
      sort((v).begin(), (v).end());
      for (long long i = 0; i < n - 1; i++) {
        if (v[i].second.first >= v[i + 1].second.first) {
          cout << v[i + 1].second.second + 1 << " " << v[i].second.second + 1;
          return 0;
        }
      }
    }
  }
  cout << -1 << " " << -1;
  return 0;
}
