#include <bits/stdc++.h>
using namespace std;
void build(int node, int l, int r, vector<pair<int, int>>& st, vector<int>& a) {
  if (l == r) {
    st[node] = {a[l], a[l]};
  } else {
    int mid = (l + r) >> 1;
    build(node << 1, l, mid, st, a);
    build(node << 1 | 1, mid + 1, r, st, a);
    st[node] = {min(st[node << 1].first, st[node << 1 | 1].first),
                max(st[node << 1].second, st[node << 1 | 1].second)};
  }
}
pair<int, int> get_tree(int node, int l, int r, int u, int v,
                        vector<pair<int, int>>& st) {
  if (v < l || r < u) return {1e9, -1};
  if (u <= l && r <= v) return st[node];
  int mid = (l + r) >> 1;
  pair<int, int> p1 = get_tree(node << 1, l, mid, u, v, st),
                 p2 = get_tree(node << 1 | 1, mid + 1, r, u, v, st);
  return {min(p1.first, p2.first), max(p1.second, p2.second)};
}
bool check(vector<pair<int, int>>& st, int l, int r, int m, int n) {
  pair<int, int> g = get_tree(1, 1, n, l, r, st);
  if (g.first == 1 && g.second == m) return true;
  return false;
}
int solve(vector<int>& a, vector<pair<int, int>>& st, vector<int>& left,
          vector<int>& right, int l, int r, int n) {
  if (r < l) return 0;
  if (l == r) return (a[l] == 1);
  int mid = (l + r) >> 1;
  int m = a[mid], b, ans = 0;
  for (int i = mid; i >= l; --i) {
    m = max(m, a[i]);
    b = i + m - 1;
    if (b >= left[i] || b <= mid || b > r) continue;
    if (check(st, i, b, m, n)) ++ans;
  }
  m = 0;
  for (int i = mid + 1; i <= r; ++i) {
    m = max(m, a[i]);
    b = i - m + 1;
    if (b <= right[i] || b > mid || b < l) continue;
    if (check(st, b, i, m, n)) ++ans;
  }
  ans += solve(a, st, left, right, l, mid, n);
  ans += solve(a, st, left, right, mid + 1, r, n);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n + 1), pos(n + 1, -1);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  set<int> s;
  vector<int> left(n + 1), right(n + 1);
  for (int i = 1, j = 1; i <= n; ++i) {
    while (j <= n && s.find(a[j]) == s.end()) {
      s.insert(a[j]);
      ++j;
    }
    left[i] = j;
    s.erase(a[i]);
  }
  s.clear();
  for (int i = n, j = n; i > 0; --i) {
    while (j > 0 && s.find(a[j]) == s.end()) {
      s.insert(a[j]);
      --j;
    }
    right[i] = j;
    s.erase(a[i]);
  }
  vector<pair<int, int>> st(4 * n + 1);
  build(1, 1, n, st, a);
  cout << solve(a, st, left, right, 1, n, n);
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

You have an array a_1, a_2, ..., a_n. 

Let's call some subarray a_l, a_{l + 1}, ... , a_r of this array a subpermutation if it contains all integers from 1 to r-l+1 exactly once. For example, array a = [2, 2, 1, 3, 2, 3, 1] contains 6 subarrays which are subpermutations: [a_2 ... a_3], [a_2 ... a_4], [a_3 ... a_3], [a_3 ... a_5], [a_5 ... a_7], [a_7 ... a_7].

You are asked to calculate the number of subpermutations.

Input

The first line contains one integer n (1 ≤ n ≤ 3 ⋅ 10^5).

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

This array can contain the same integers.

Output

Print the number of subpermutations of the array a.

Examples

Input


8
2 4 1 3 4 2 1 2


Output


7


Input


5
1 1 2 1 2


Output


6

Note

There are 7 subpermutations in the first test case. Their segments of indices are [1, 4], [3, 3], [3, 6], [4, 7], [6, 7], [7, 7] and [7, 8].

In the second test case 6 subpermutations exist: [1, 1], [2, 2], [2, 3], [3, 4], [4, 4] and [4, 5].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void build(int node, int l, int r, vector<pair<int, int>>& st, vector<int>& a) {
  if (l == r) {
    st[node] = {a[l], a[l]};
  } else {
    int mid = (l + r) >> 1;
    build(node << 1, l, mid, st, a);
    build(node << 1 | 1, mid + 1, r, st, a);
    st[node] = {min(st[node << 1].first, st[node << 1 | 1].first),
                max(st[node << 1].second, st[node << 1 | 1].second)};
  }
}
pair<int, int> get_tree(int node, int l, int r, int u, int v,
                        vector<pair<int, int>>& st) {
  if (v < l || r < u) return {1e9, -1};
  if (u <= l && r <= v) return st[node];
  int mid = (l + r) >> 1;
  pair<int, int> p1 = get_tree(node << 1, l, mid, u, v, st),
                 p2 = get_tree(node << 1 | 1, mid + 1, r, u, v, st);
  return {min(p1.first, p2.first), max(p1.second, p2.second)};
}
bool check(vector<pair<int, int>>& st, int l, int r, int m, int n) {
  pair<int, int> g = get_tree(1, 1, n, l, r, st);
  if (g.first == 1 && g.second == m) return true;
  return false;
}
int solve(vector<int>& a, vector<pair<int, int>>& st, vector<int>& left,
          vector<int>& right, int l, int r, int n) {
  if (r < l) return 0;
  if (l == r) return (a[l] == 1);
  int mid = (l + r) >> 1;
  int m = a[mid], b, ans = 0;
  for (int i = mid; i >= l; --i) {
    m = max(m, a[i]);
    b = i + m - 1;
    if (b >= left[i] || b <= mid || b > r) continue;
    if (check(st, i, b, m, n)) ++ans;
  }
  m = 0;
  for (int i = mid + 1; i <= r; ++i) {
    m = max(m, a[i]);
    b = i - m + 1;
    if (b <= right[i] || b > mid || b < l) continue;
    if (check(st, b, i, m, n)) ++ans;
  }
  ans += solve(a, st, left, right, l, mid, n);
  ans += solve(a, st, left, right, mid + 1, r, n);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n + 1), pos(n + 1, -1);
  for (int i = 1; i <= n; ++i) cin >> a[i];
  set<int> s;
  vector<int> left(n + 1), right(n + 1);
  for (int i = 1, j = 1; i <= n; ++i) {
    while (j <= n && s.find(a[j]) == s.end()) {
      s.insert(a[j]);
      ++j;
    }
    left[i] = j;
    s.erase(a[i]);
  }
  s.clear();
  for (int i = n, j = n; i > 0; --i) {
    while (j > 0 && s.find(a[j]) == s.end()) {
      s.insert(a[j]);
      --j;
    }
    right[i] = j;
    s.erase(a[i]);
  }
  vector<pair<int, int>> st(4 * n + 1);
  build(1, 1, n, st, a);
  cout << solve(a, st, left, right, 1, n, n);
  return 0;
}

```