#include <bits/stdc++.h>
using namespace std;
const int siz = 1e3;
int n, arr[siz + 5];
map<long long, int> nb;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
  }
  long long open = 0, ans = 0;
  int lb = 0;
  for (int i = 1; i <= n; ++i) {
    if (i & 1) {
      nb[open] = lb;
      open += arr[i];
    } else {
      int cur = arr[i];
      lb = 0;
      while (!nb.empty()) {
        auto it = nb.rbegin();
        if (open - it->first <= cur) {
          long long sub = open - it->first;
          ans += it->second + sub;
          cur -= sub;
          open -= sub;
          lb = it->second + 1;
          nb.erase(it->first);
        } else {
          break;
        }
      }
      if (cur) {
        long long sub = min(open, (long long)cur);
        ans += sub;
        open -= sub;
        cur -= sub;
        lb = !cur;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
const int siz = 1e3;
int n, arr[siz + 5];
map<long long, int> nb;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &arr[i]);
  }
  long long open = 0, ans = 0;
  int lb = 0;
  for (int i = 1; i <= n; ++i) {
    if (i & 1) {
      nb[open] = lb;
      open += arr[i];
    } else {
      int cur = arr[i];
      lb = 0;
      while (!nb.empty()) {
        auto it = nb.rbegin();
        if (open - it->first <= cur) {
          long long sub = open - it->first;
          ans += it->second + sub;
          cur -= sub;
          open -= sub;
          lb = it->second + 1;
          nb.erase(it->first);
        } else {
          break;
        }
      }
      if (cur) {
        long long sub = min(open, (long long)cur);
        ans += sub;
        open -= sub;
        cur -= sub;
        lb = !cur;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```