#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
long long n, a[300010], rep[300010], pre[300010], sta[300010], ans = 0, rd = 0;
struct RandGen {
  long long x, y, z;
  long long nextInt() {
    long long t = x ^ (x << 11);
    x = y;
    y = z;
    z ^= (z >> 19) ^ t ^ (t >> 8);
    return x;
  }
  long long random() { return nextInt() % (long long)937183548295; }
} rng;
bool check(long long l, long long r) {
  long long ck;
  if (l == 0) {
    ck = pre[r];
  } else {
    ck = pre[r] ^ pre[l - 1];
  }
  if (ck == sta[r - l + 1]) {
    return true;
  }
  return false;
}
int main() {
  long long i, j, mx;
  rng.x = 18284;
  rng.y = 175393;
  rng.z = 10573223;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  for (i = 1; i <= n; i++) {
    rep[i] = rng.random();
    sta[i] = sta[i - 1] ^ rep[i];
  }
  pre[0] = rep[a[0]];
  for (i = 1; i < n; i++) {
    pre[i] = pre[i - 1] ^ rep[a[i]];
  }
  for (i = 0; i < n; i++) {
    if (a[i] == 1) {
      mx = 1;
      for (j = i; j >= 0; j--) {
        if (a[j] == 1 && j != i) {
          break;
        }
        mx = max(mx, a[j]);
        if (check(j, j + mx - 1)) {
          ans++;
        }
      }
      mx = 1;
      for (j = i + 1; j < n; j++) {
        if (a[j] == 1) {
          break;
        }
        mx = max(mx, a[j]);
        if (check(j - mx + 1, j)) {
          ans++;
        }
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize(2)
using namespace std;
long long n, a[300010], rep[300010], pre[300010], sta[300010], ans = 0, rd = 0;
struct RandGen {
  long long x, y, z;
  long long nextInt() {
    long long t = x ^ (x << 11);
    x = y;
    y = z;
    z ^= (z >> 19) ^ t ^ (t >> 8);
    return x;
  }
  long long random() { return nextInt() % (long long)937183548295; }
} rng;
bool check(long long l, long long r) {
  long long ck;
  if (l == 0) {
    ck = pre[r];
  } else {
    ck = pre[r] ^ pre[l - 1];
  }
  if (ck == sta[r - l + 1]) {
    return true;
  }
  return false;
}
int main() {
  long long i, j, mx;
  rng.x = 18284;
  rng.y = 175393;
  rng.z = 10573223;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  for (i = 1; i <= n; i++) {
    rep[i] = rng.random();
    sta[i] = sta[i - 1] ^ rep[i];
  }
  pre[0] = rep[a[0]];
  for (i = 1; i < n; i++) {
    pre[i] = pre[i - 1] ^ rep[a[i]];
  }
  for (i = 0; i < n; i++) {
    if (a[i] == 1) {
      mx = 1;
      for (j = i; j >= 0; j--) {
        if (a[j] == 1 && j != i) {
          break;
        }
        mx = max(mx, a[j]);
        if (check(j, j + mx - 1)) {
          ans++;
        }
      }
      mx = 1;
      for (j = i + 1; j < n; j++) {
        if (a[j] == 1) {
          break;
        }
        mx = max(mx, a[j]);
        if (check(j - mx + 1, j)) {
          ans++;
        }
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```