#include <bits/stdc++.h>
using namespace std;
const int Nmax = 1010;
long long minv[Nmax][Nmax], s[Nmax];
long long a[Nmax];
int n;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + (i % 2 == 1 ? 1 : -1) * a[i];
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    minv[i][i - 1] = 0;
    long long now = 0;
    for (int j = i; j <= n; ++j) {
      now += (j % 2 == 1 ? 1 : -1) * a[j];
      minv[i][j] = min(minv[i][j - 1], now);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      long long lv = minv[i + 1][j - 1];
      if (i % 2 == 0 || j % 2 == 1) continue;
      long long l = a[i] + lv;
      long long r = a[j] - (s[j - 1] - s[i] - lv);
      int okl = (lv != 0), okr = (s[j - 1] - s[i] - lv != 0);
      ans += max(0LL, min(l + okl, r + okr));
    }
  }
  cout << ans << "\n";
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _ = 1;
  while (_--) {
    solve();
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

<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 Nmax = 1010;
long long minv[Nmax][Nmax], s[Nmax];
long long a[Nmax];
int n;
void solve() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + (i % 2 == 1 ? 1 : -1) * a[i];
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    minv[i][i - 1] = 0;
    long long now = 0;
    for (int j = i; j <= n; ++j) {
      now += (j % 2 == 1 ? 1 : -1) * a[j];
      minv[i][j] = min(minv[i][j - 1], now);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = i + 1; j <= n; ++j) {
      long long lv = minv[i + 1][j - 1];
      if (i % 2 == 0 || j % 2 == 1) continue;
      long long l = a[i] + lv;
      long long r = a[j] - (s[j - 1] - s[i] - lv);
      int okl = (lv != 0), okr = (s[j - 1] - s[i] - lv != 0);
      ans += max(0LL, min(l + okl, r + okr));
    }
  }
  cout << ans << "\n";
}
int main() {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _ = 1;
  while (_--) {
    solve();
  }
}

```