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>
const int mod = 1e9 + 7;
const int maxn = 3e5 + 5;
const double eps = 1e-6;
using namespace std;
bool eq(const double &a, const double &b) { return fabs(a - b) < eps; }
bool ls(const double &a, const double &b) { return a + eps < b; }
bool le(const double &a, const double &b) { return eq(a, b) || ls(a, b); }
long long gcd(long long a, long long b) { return a == 0 ? b : gcd(b % a, a); };
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long kpow(long long a, long long b) {
  long long res = 1;
  a %= mod;
  if (b < 0) return 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
struct edge {
  int l, r, id;
  bool operator<(edge c) const {
    if (l == c.l) return r > c.r;
    return l < c.l;
  }
} e[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (n); ++i) {
    scanf("%d%d", &e[i].l, &e[i].r);
    e[i].id = i + 1;
  }
  sort(e, e + n);
  int i = 0;
  while (i < n) {
    int j = i + 1;
    while (1) {
      if (e[j].r > e[i].r || j > n) break;
      if (e[j].l >= e[i].l && e[j].r <= e[i].r)
        return printf("%d %d\n", e[j].id, e[i].id), 0;
      ++j;
    }
    i = j;
  }
  puts("-1 -1");
  return 0;
}
