#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
long long int c[N], pref[N];
long long int sum(int i, int j) {
  long long int ans = pref[j];
  if (i) ans -= pref[i - 1];
  return ans;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    if (i) pref[i] = pref[i - 1];
    if (i % 2 == 0)
      pref[i] += c[i];
    else
      pref[i] -= c[i];
  }
  long long int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (i & 1) continue;
    long long int prefix = 0, k = 0;
    for (int j = i + 1; j < n; ++j) {
      if (j & 1) {
        if (c[i] >= k) {
          long long int sum = prefix + k;
          if (c[j] >= sum && sum > 0) ++ans;
          ans += max(0LL, min(c[i] - k, c[j] - max(0LL, sum)));
        }
        prefix -= c[j];
      } else
        prefix += c[j];
      if (prefix < 0) k = max(k, -prefix);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

### 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;
const int N = 1005;
long long int c[N], pref[N];
long long int sum(int i, int j) {
  long long int ans = pref[j];
  if (i) ans -= pref[i - 1];
  return ans;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> c[i];
    if (i) pref[i] = pref[i - 1];
    if (i % 2 == 0)
      pref[i] += c[i];
    else
      pref[i] -= c[i];
  }
  long long int ans = 0;
  for (int i = 0; i < n; ++i) {
    if (i & 1) continue;
    long long int prefix = 0, k = 0;
    for (int j = i + 1; j < n; ++j) {
      if (j & 1) {
        if (c[i] >= k) {
          long long int sum = prefix + k;
          if (c[j] >= sum && sum > 0) ++ans;
          ans += max(0LL, min(c[i] - k, c[j] - max(0LL, sum)));
        }
        prefix -= c[j];
      } else
        prefix += c[j];
      if (prefix < 0) k = max(k, -prefix);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}

```