#include <bits/stdc++.h>
using namespace std;
template <class... Args>
void debug(Args... args) {
  auto tmp = {(cout << args << '\t', 0)...};
  cout << endl;
}
using ll = long long;
const int N = 1e3 + 10;
const int mod = 4933;
const double eps = 1e-9;
ll a[N];
int n;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  ll ans = 0;
  for (int i = 1; i <= n; i += 2) {
    ll x = 0;
    ll y = 0;
    ll z = 1;
    for (int j = i + 1; j <= n; j += 2) {
      ll d = abs(x - y);
      if (x == y) {
        ans += max(0ll, min(a[i], a[j]) - z + 1);
      } else if (x > y) {
        ans += max(0ll, min(a[i], a[j] - d) - z + 1);
      } else {
        ans += max(0ll, min(a[i] - d, a[j]) - max(1ll, z - d) + 1);
      }
      y += a[j];
      z = max(z, y - x);
      x += a[j + 1];
    }
  }
  cout << ans;
}

### Prompt

Develop a solution in Cpp 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;
template <class... Args>
void debug(Args... args) {
  auto tmp = {(cout << args << '\t', 0)...};
  cout << endl;
}
using ll = long long;
const int N = 1e3 + 10;
const int mod = 4933;
const double eps = 1e-9;
ll a[N];
int n;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  ll ans = 0;
  for (int i = 1; i <= n; i += 2) {
    ll x = 0;
    ll y = 0;
    ll z = 1;
    for (int j = i + 1; j <= n; j += 2) {
      ll d = abs(x - y);
      if (x == y) {
        ans += max(0ll, min(a[i], a[j]) - z + 1);
      } else if (x > y) {
        ans += max(0ll, min(a[i], a[j] - d) - z + 1);
      } else {
        ans += max(0ll, min(a[i] - d, a[j]) - max(1ll, z - d) + 1);
      }
      y += a[j];
      z = max(z, y - x);
      x += a[j + 1];
    }
  }
  cout << ans;
}

```