#include <bits/stdc++.h>
using namespace std;
long long intersect_len(pair<long long, long long> A,
                        pair<long long, long long> B) {
  const auto &[la, ra] = A;
  const auto &[lb, rb] = B;
  if (ra < lb || rb < la) {
    return 0;
  }
  return min(ra, rb) - max(la, lb) + 1;
}
int main(void) {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> c(n + 1);
  for (int i = 0; i < (int)(n); i++) {
    cin >> c[i + 1];
  }
  long long res = 0;
  for (int l = 1; l <= n; l += 2) {
    long long mn = 0, tot = 0;
    for (int r = l + 1; r <= n; r += 2) {
      if (c[l] >= -mn) {
        res += intersect_len({max(1ll, -mn), c[l]}, {1 - tot, c[r] - tot});
      }
      tot -= c[r];
      mn = min(mn, tot);
      if (r + 1 <= n) {
        tot += c[r + 1];
      }
    }
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

<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;
long long intersect_len(pair<long long, long long> A,
                        pair<long long, long long> B) {
  const auto &[la, ra] = A;
  const auto &[lb, rb] = B;
  if (ra < lb || rb < la) {
    return 0;
  }
  return min(ra, rb) - max(la, lb) + 1;
}
int main(void) {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> c(n + 1);
  for (int i = 0; i < (int)(n); i++) {
    cin >> c[i + 1];
  }
  long long res = 0;
  for (int l = 1; l <= n; l += 2) {
    long long mn = 0, tot = 0;
    for (int r = l + 1; r <= n; r += 2) {
      if (c[l] >= -mn) {
        res += intersect_len({max(1ll, -mn), c[l]}, {1 - tot, c[r] - tot});
      }
      tot -= c[r];
      mn = min(mn, tot);
      if (r + 1 <= n) {
        tot += c[r + 1];
      }
    }
  }
  cout << res << "\n";
  return 0;
}

```