#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 100;
int n, a[N], ans;
mt19937 rnd(1);
unsigned long long val[N], Xor[N], sum[N];
bool check(int l, int r) {
  if (l < 0 || r > n || l > r) return 0;
  return !(sum[r] ^ sum[l - 1] ^ Xor[r - l + 1]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i], val[i] = rnd();
  for (int i = 1; i <= n; ++i) Xor[i] = Xor[i - 1] ^ val[i];
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] ^ val[a[i]];
  for (int i = 1; i <= n; ++i)
    if (a[i] == 1) {
      int l = i, r = i, mxl = 0, mxr = 0;
      ++ans;
      while (l > 1 && a[l - 1] != 1)
        mxl = max(mxl, a[--l]), ans += check(l, l + mxl - 1);
      while (r < n && a[r + 1] != 1)
        mxr = max(mxr, a[++r]), ans += check(r - mxr + 1, r);
    }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 int N = 3e5 + 100;
int n, a[N], ans;
mt19937 rnd(1);
unsigned long long val[N], Xor[N], sum[N];
bool check(int l, int r) {
  if (l < 0 || r > n || l > r) return 0;
  return !(sum[r] ^ sum[l - 1] ^ Xor[r - l + 1]);
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i], val[i] = rnd();
  for (int i = 1; i <= n; ++i) Xor[i] = Xor[i - 1] ^ val[i];
  for (int i = 1; i <= n; ++i) sum[i] = sum[i - 1] ^ val[a[i]];
  for (int i = 1; i <= n; ++i)
    if (a[i] == 1) {
      int l = i, r = i, mxl = 0, mxr = 0;
      ++ans;
      while (l > 1 && a[l - 1] != 1)
        mxl = max(mxl, a[--l]), ans += check(l, l + mxl - 1);
      while (r < n && a[r + 1] != 1)
        mxr = max(mxr, a[++r]), ans += check(r - mxr + 1, r);
    }
  cout << ans << endl;
  return 0;
}

```