n = int(input())
l, r = 1, n
a = list(map(int, input().split()))
dd = dict()
for idx, i in enumerate(a):
try:
dd[i] = (dd[i][0]+1, dd[i][1], idx+1, max(dd[i][3], idx+1 - dd[i][1]))
except:
dd[i] = (1, idx+1, idx+1, 0)
d = n
mx = max(dd.values())[0]
temp = [i for i in dd.values() if i[0] == mx]
m_d = sorted(temp, key=lambda x: x[3])
temp_1 = [i for i in temp if i[3] == m_d[0][3]]
_, l, r, _ = temp_1[0]
print(l, r)
#include "bits/stdc++.h"
using namespace std;
#define ff first
#define ss second
void solve();
#pragma GCC optimize("Ofast","inline","-ffast-math")
const int N = 101000;
int n, mx, d[N], dk[N * 10], firs[N * 10], last[N * 10];
int main(){
// int t = 1; cin >> t;
// while(t-- > 0) solve();
cin >> n;
for(int i = 1; i <= n; i++){
cin >> d[i];
dk[d[i]]++;
mx = max(mx, dk[d[i]]);
if(!firs[d[i]]) firs[d[i]] = i;
last[d[i]] = i;
} int res = N, cur1, cur2;
for(int i = 1; i < N * 10; i++){
if(dk[i] == mx)
if(res > last[i] - firs[i] + 1){
res = last[i] - firs[i] + 1;
cur1 = firs[i]; cur2 = last[i];
}
} cout << cur1 << " " << cur2;
}
const int fastIO = [](){
ios_base::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);
return 0;
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