You are given an array a consisting of n integers. We denote the subarray a[l..r] as the array [a_l, a_{l + 1}, ..., a_r] (1 ≤ l ≤ r ≤ n).

A subarray is considered good if every integer that occurs in this subarray occurs there exactly thrice. For example, the array [1, 2, 2, 2, 1, 1, 2, 2, 2] has three good subarrays:

  * a[1..6] = [1, 2, 2, 2, 1, 1]; 
  * a[2..4] = [2, 2, 2]; 
  * a[7..9] = [2, 2, 2]. 



Calculate the number of good subarrays of the given array a.

Input

The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

Output

Print one integer — the number of good subarrays of the array a.

Examples

Input


9
1 2 2 2 1 1 2 2 2


Output


3


Input


10
1 2 3 4 1 2 3 1 2 3


Output


0


Input


12
1 2 3 4 3 4 2 1 3 4 2 1


Output


1
#include <bits/stdc++.h>
using namespace std;
const long long int INF = 9223372036854775807;
const long long int mod = (long long int)998244353;
#pragma GCC target("avx2")
#pragma GCC optimization("unroll-loops")
int MOD(int a, int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
long long int max3(long long int a, long long int b, long long int c) {
  return max(c, max(a, b));
}
long long int min3(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
long long int power(long long int x, long long int y, long long int mod1) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod1;
    y = y >> 1;
    x = (x * x) % mod1;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int abso(long long int x) {
  if (x < 0) {
    return -x;
  }
  return x;
}
long long int ceiling(long long int a, long long int b) {
  if (a % b == 0) {
    return a / b;
  } else {
    return a / b + 1;
  }
}
long long int modinv(long long int x, long long int p) {
  return power(x, p - 2, p);
}
struct point {
  long long int x;
  long long int y;
};
long long int cross(point a, point b) { return a.x * b.y - b.x * a.y; }
bool ccw(point p1, point p2) {
  if (cross(p1, p2) > 0) {
    return true;
  } else {
    return false;
  }
}
void ccw_sort(vector<point> &v) {
  vector<point> v1, v2;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].y > 0) {
      v1.push_back(v[i]);
    } else {
      v2.push_back(v[i]);
    }
  }
  sort(v1.begin(), v1.end(), ccw);
  sort(v2.begin(), v2.end(), ccw);
  for (int i = 0; i < v2.size(); i++) {
    v1.push_back(v2[i]);
  }
  v = v1;
}
void dfs(int i, int p, vector<int> &sz, vector<vector<int> > &adj,
         vector<int> &p1) {
  sz[i] = 1;
  p1[i] = p;
  for (int j = 0; j < adj[i].size(); j++) {
    if (adj[i][j] != p) {
      dfs(adj[i][j], i, sz, adj, p1);
      sz[i] = sz[i] + sz[adj[i][j]];
    }
  }
}
vector<long long int> fact, ifact;
long long int func(long long int n, long long int r) {
  if (n < r) {
    return 0;
  } else {
    return ((fact[n] * ((ifact[n - r] * ifact[r]) % mod)) % mod);
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int b = 5449;
  long long int bb = 7919;
  long long int mod1 = 1e9 + 7;
  vector<vector<int> > stk(n + 1);
  long long int hash1 = 0;
  long long int hash2 = 0;
  vector<int> cnt(n + 1);
  set<int> g;
  map<pair<long long int, long long int>, vector<int> > g2;
  long long int ans = 0;
  g2[make_pair(0, 0)].push_back(-1);
  for (int i = 0; i < n; i++) {
    stk[a[i]].push_back(i);
    hash1 = hash1 - (((cnt[a[i]]) * power(b, a[i] + 1, mod)) % mod);
    hash1 = (hash1 + mod);
    hash1 = hash1 % mod;
    hash2 = hash2 - (((cnt[a[i]]) * power(bb, a[i] + 1, mod1)) % mod1);
    hash2 = (hash2 + mod1);
    hash2 = hash2 % mod1;
    cnt[a[i]]++;
    cnt[a[i]] = cnt[a[i]] % 3;
    hash1 = hash1 + (((cnt[a[i]]) * power(b, a[i] + 1, mod)) % mod);
    hash1 = hash1 % mod;
    hash2 = hash2 + (((cnt[a[i]]) * power(bb, a[i] + 1, mod1)) % mod1);
    hash2 = hash2 % mod1;
    g2[make_pair(hash1, hash2)].push_back(i);
    if (stk[a[i]].size() > 3) {
      g.insert(stk[a[i]][0]);
      stk[a[i]].erase(stk[a[i]].begin());
    }
    if (g.size() > 0) {
      ans = ans +
            (g2[make_pair(hash1, hash2)].end() -
             lower_bound(g2[make_pair(hash1, hash2)].begin(),
                         g2[make_pair(hash1, hash2)].end(), *g.rbegin())) -
            1;
    } else {
      ans = ans +
            (g2[make_pair(hash1, hash2)].end() -
             lower_bound(g2[make_pair(hash1, hash2)].begin(),
                         g2[make_pair(hash1, hash2)].end(), -1)) -
            1;
    }
  }
  cout << ans << endl;
}
