#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize ("Ofast")
#define F first
#define S second
#define vi vector<int>
#define vvi vector<vi>
#define pi pair<int, int>
#define vpi vector<pi>
#define vb vector<bool>
#define vvb vector<vb>
#define pb push_back
#define ppb pop_back
#define read(a) for(auto &x:a) cin >> x;
#define print(a) for(auto x:a) cout << x << " "; cout << "\n";
#define vc vector<char>
#define vvc vector<vc>
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define int long long
#define ld long double
const int INF = 1e18;
const int inf = 1e9;
struct FT{
int n;
vi tree;
FT(int sz){
n = sz;
tree = vi(n+1);
}
void update(int k, int val){
k++;
while(k <= n){
tree[k] += val;
k += k&-k;
}
}
int query(int l, int r){
r++;
int ans = 0;
while(r >= 1){
ans += tree[r];
r -= r&-r;
}
while(l >= 1){
ans -= tree[l];
l -= l&-l;
}
return ans;
}
};
void solve(){
int n; cin >> n;
vi a(n); read(a);
map<int, vi> mp;
for(int i=0; i<n; i++)
mp[a[i]].pb(i);
//Número de subarrays de longitud <= len en [l, r]
auto f = [&](int l, int r, int len){
if(l > r) return 0LL;
int s = (r-l+1)*(r-l+2)/2;
if(len < r-l+1) s -= (r-l+1-len)*(r-l+2-len)/2;
return s;
};
int ans = 0;
for(auto &[el, pos]:mp){
if(pos.size() <= 1000){
for(int i=0; i<pos.size(); i++)
for(int j=i; j<pos.size(); j++){
int mxlen = 2*(j-i)+1;
if(pos[j]-pos[i]+1 > mxlen) continue;
int l = (i ? pos[i-1]+1 : 0);
int r = (j+1 < pos.size() ? pos[j+1]-1 : n-1);
ans += f(l, r, mxlen)-f(pos[i]+1, r, mxlen)-f(l, pos[j]-1, mxlen)+
f(pos[i]+1, pos[j]-1, mxlen);
}
} else {
FT ft(2*n+1);
int curr = 0;
ft.update(n, 1);
for(int i=0; i<n; i++){
curr += (a[i] == el ? 1 : -1);
if(curr+n-1 >= 0) ans += ft.query(0, curr+n-1);
ft.update(curr+n, 1);
}
}
}
cout << ans << "\n";
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
// #ifndef ONLINE_JUDGE
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// #endif
int tt = 1;
// cin >> tt;
while(tt--)
solve();
return 0;
}
