#include <bits/stdc++.h>
using namespace std;
inline int read() {
  register int s = 0, f = 0;
  register char ch = getchar();
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 1) + (s << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -s : s;
}
inline void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
long long a[1005], n, ans;
signed main() {
  cin >> n;
  for (auto i = 1; i <= n; i += 1) {
    cin >> a[i];
  }
  for (auto i = 1; i <= n; i += 2) {
    long long sum = 0;
    long long mn = a[i] - 1;
    for (auto j = i + 1; j <= n; j += 2) {
      sum += a[j - 1];
      sum -= a[j];
      if (sum < 0) {
        ans += min(a[j] + sum + 1, mn + 1);
        break;
      } else {
        ans += min(a[j], max(0LL, mn - sum + 1));
        if (j == i + 1) {
          mn = min(a[i], sum);
        } else {
          mn = min(mn, sum);
        }
      }
    }
  }
  cout << 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;
inline int read() {
  register int s = 0, f = 0;
  register char ch = getchar();
  while (!isdigit(ch)) {
    f |= (ch == '-');
    ch = getchar();
  }
  while (isdigit(ch)) {
    s = (s << 1) + (s << 3) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -s : s;
}
inline void write(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
long long a[1005], n, ans;
signed main() {
  cin >> n;
  for (auto i = 1; i <= n; i += 1) {
    cin >> a[i];
  }
  for (auto i = 1; i <= n; i += 2) {
    long long sum = 0;
    long long mn = a[i] - 1;
    for (auto j = i + 1; j <= n; j += 2) {
      sum += a[j - 1];
      sum -= a[j];
      if (sum < 0) {
        ans += min(a[j] + sum + 1, mn + 1);
        break;
      } else {
        ans += min(a[j], max(0LL, mn - sum + 1));
        if (j == i + 1) {
          mn = min(a[i], sum);
        } else {
          mn = min(mn, sum);
        }
      }
    }
  }
  cout << ans;
  return 0;
}

```