#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
long long a[300005], b[300005], c[300005], d[300005];
int32_t main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0, j = 0; i < n; i++) {
    if (i) c[a[i - 1]]--;
    while (j < n && c[a[j]] == 0) c[a[j++]]++;
    b[i] = j - 1;
  }
  d[0] = a[0];
  for (long long i = 1; i < n; i++) d[i] = d[i - 1] + a[i];
  vector<long long> v({-1});
  for (long long i = 0; i < n; i++)
    if (a[i] == 1) v.push_back(i);
  v.push_back(n);
  long long ans = 0;
  for (long long i = 1; i < v.size() - 1; i++) {
    long long r = v[i + 1] - 1;
    long long l = v[i - 1] + 1;
    long long mx = -87;
    for (long long j = v[i]; j <= r; j++) {
      mx = max(mx, a[j]);
      long long k = j - mx + 1;
      if (k >= 0 && b[k] >= j && d[j] - d[k - 1] == mx * (mx + 1) / 2) ans++;
    }
    mx = -87;
    for (long long j = v[i] - 1; j >= l; j--) {
      mx = max(mx, a[j]);
      long long k = j + mx - 1;
      if (k < n && b[j] >= k && d[k] - d[j - 1] == mx * (mx + 1) / 2) ans++;
    }
  }
  cout << ans << endl;
}

### 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;
const long long INF = 1e9 + 7;
long long a[300005], b[300005], c[300005], d[300005];
int32_t main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0, j = 0; i < n; i++) {
    if (i) c[a[i - 1]]--;
    while (j < n && c[a[j]] == 0) c[a[j++]]++;
    b[i] = j - 1;
  }
  d[0] = a[0];
  for (long long i = 1; i < n; i++) d[i] = d[i - 1] + a[i];
  vector<long long> v({-1});
  for (long long i = 0; i < n; i++)
    if (a[i] == 1) v.push_back(i);
  v.push_back(n);
  long long ans = 0;
  for (long long i = 1; i < v.size() - 1; i++) {
    long long r = v[i + 1] - 1;
    long long l = v[i - 1] + 1;
    long long mx = -87;
    for (long long j = v[i]; j <= r; j++) {
      mx = max(mx, a[j]);
      long long k = j - mx + 1;
      if (k >= 0 && b[k] >= j && d[j] - d[k - 1] == mx * (mx + 1) / 2) ans++;
    }
    mx = -87;
    for (long long j = v[i] - 1; j >= l; j--) {
      mx = max(mx, a[j]);
      long long k = j + mx - 1;
      if (k < n && b[j] >= k && d[k] - d[j - 1] == mx * (mx + 1) / 2) ans++;
    }
  }
  cout << ans << endl;
}

```