#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n, 0);
  for (long long &i : v) {
    cin >> i;
  }
  stack<pair<long long, long long>> s;
  s.push({0, 0});
  long long depth = 0;
  long long total = 0;
  for (int i = 0; i + 1 < n; i += 2) {
    long long a = v[i];
    long long b = v[i + 1];
    long long new_depth = depth + a - b;
    total += min(b, depth + a);
    while (!s.empty() && new_depth < s.top().first) {
      total += s.top().second;
      s.pop();
    }
    if (!s.empty() && s.top().first == new_depth) {
      total += s.top().second;
      s.top().second++;
    } else {
      s.push({new_depth, 1});
    }
    depth = new_depth;
    if (depth < 0) {
      depth = 0;
      s = stack<pair<long long, long long>>();
    }
  }
  cout << total << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

<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;
void solve() {
  int n;
  cin >> n;
  vector<long long> v(n, 0);
  for (long long &i : v) {
    cin >> i;
  }
  stack<pair<long long, long long>> s;
  s.push({0, 0});
  long long depth = 0;
  long long total = 0;
  for (int i = 0; i + 1 < n; i += 2) {
    long long a = v[i];
    long long b = v[i + 1];
    long long new_depth = depth + a - b;
    total += min(b, depth + a);
    while (!s.empty() && new_depth < s.top().first) {
      total += s.top().second;
      s.pop();
    }
    if (!s.empty() && s.top().first == new_depth) {
      total += s.top().second;
      s.top().second++;
    } else {
      s.push({new_depth, 1});
    }
    depth = new_depth;
    if (depth < 0) {
      depth = 0;
      s = stack<pair<long long, long long>>();
    }
  }
  cout << total << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
}

```