#include <bits/stdc++.h>
using namespace std;
long long ans, f[1005][1005], a[1005], n, sub, res;
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    f[i][i] = f[i - 1][i - 1] + (i & 1 ? a[i] : -a[i]);
  }
  for (int len = 2; len <= n; ++len)
    for (int i = 1; i + len - 1 <= n; ++i)
      f[i][i + len - 1] = min(f[i][i + len - 2], f[i + 1][i + len - 1]);
  for (int i = 1; i <= n; ++i)
    if (i & 1)
      for (int j = i + 1; j <= n; j += 2) {
        sub = j >= i + 2 ? f[i][i] - f[i + 1][j - 1] : 0;
        res = min(a[i] - sub, a[j] - sub - f[j - 1][j - 1] + f[i][i]);
        if (res >= 0) ans += res + (j >= i + 2);
      }
  printf("%lld", ans);
  return 0;
}

### Prompt

Generate 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 ans, f[1005][1005], a[1005], n, sub, res;
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    f[i][i] = f[i - 1][i - 1] + (i & 1 ? a[i] : -a[i]);
  }
  for (int len = 2; len <= n; ++len)
    for (int i = 1; i + len - 1 <= n; ++i)
      f[i][i + len - 1] = min(f[i][i + len - 2], f[i + 1][i + len - 1]);
  for (int i = 1; i <= n; ++i)
    if (i & 1)
      for (int j = i + 1; j <= n; j += 2) {
        sub = j >= i + 2 ? f[i][i] - f[i + 1][j - 1] : 0;
        res = min(a[i] - sub, a[j] - sub - f[j - 1][j - 1] + f[i][i]);
        if (res >= 0) ans += res + (j >= i + 2);
      }
  printf("%lld", ans);
  return 0;
}

```