#include <bits/stdc++.h>
using namespace std;
const long long int INF{(false ? (INT_MAX - 1) / 2 : (long long int)1e18 + 5)};
long long int MOD = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      long long int l = a[i];
      if (i + 1 < n) {
        long long int r = min(l, a[i + 1]);
        ans += r;
        if (a[i + 1] - r > 0)
          continue;
        else {
          l -= r;
        }
      }
      long long int p = 0, p1 = 0, q = 0;
      for (long long int j = i + 2; j < n; j++) {
        if (j % 2 == 0)
          p += a[j], p1 = p - a[j];
        else {
          q = a[j];
          long long int r = min(q, p);
          p -= r;
          q -= r;
          if (p == 0) {
            ans += 1;
            long long int m = min(l, q);
            ans += m;
            l -= m;
            q -= m;
            if (q > 0) break;
          }
        }
      }
    }
  }
  cout << ans;
}
int main() {
  cout << setprecision(15);
  long long int t(1);
  while (t--) {
    solve();
    cout << endl;
  }
  return 0;
}

### Prompt

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

<image>

William has a favorite bracket sequence. Since his favorite sequence is quite big he provided it to you as a sequence of positive integers c_1, c_2, ..., c_n where c_i is the number of consecutive brackets "(" if i is an odd number or the number of consecutive brackets ")" if i is an even number.

For example for a bracket sequence "((())()))" a corresponding sequence of numbers is [3, 2, 1, 3].

You need to find the total number of continuous subsequences (subsegments) [l, r] (l ≤ r) of the original bracket sequence, which are regular bracket sequences.

A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters "+" and "1" into this sequence. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not.

Input

The first line contains a single integer n (1 ≤ n ≤ 1000), the size of the compressed sequence.

The second line contains a sequence of integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ 10^9), the compressed sequence.

Output

Output a single integer — the total number of subsegments of the original bracket sequence, which are regular bracket sequences.

It can be proved that the answer fits in the signed 64-bit integer data type.

Examples

Input


5
4 1 2 3 1


Output


5


Input


6
1 3 2 1 2 4


Output


6


Input


6
1 1 1 1 2 2


Output


7

Note

In the first example a sequence (((()(()))( is described. This bracket sequence contains 5 subsegments which form regular bracket sequences:

  1. Subsequence from the 3rd to 10th character: (()(()))
  2. Subsequence from the 4th to 5th character: ()
  3. Subsequence from the 4th to 9th character: ()(())
  4. Subsequence from the 6th to 9th character: (())
  5. Subsequence from the 7th to 8th character: ()



In the second example a sequence ()))(()(()))) is described.

In the third example a sequence ()()(()) is described.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long int INF{(false ? (INT_MAX - 1) / 2 : (long long int)1e18 + 5)};
long long int MOD = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      long long int l = a[i];
      if (i + 1 < n) {
        long long int r = min(l, a[i + 1]);
        ans += r;
        if (a[i + 1] - r > 0)
          continue;
        else {
          l -= r;
        }
      }
      long long int p = 0, p1 = 0, q = 0;
      for (long long int j = i + 2; j < n; j++) {
        if (j % 2 == 0)
          p += a[j], p1 = p - a[j];
        else {
          q = a[j];
          long long int r = min(q, p);
          p -= r;
          q -= r;
          if (p == 0) {
            ans += 1;
            long long int m = min(l, q);
            ans += m;
            l -= m;
            q -= m;
            if (q > 0) break;
          }
        }
      }
    }
  }
  cout << ans;
}
int main() {
  cout << setprecision(15);
  long long int t(1);
  while (t--) {
    solve();
    cout << endl;
  }
  return 0;
}

```