#include <bits/stdc++.h>
using namespace std;
const signed long long MOD = 1e9 + 7;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
signed long long n;
signed long long c[1001];
pair<signed long long, signed long long> intersect(
    pair<signed long long, signed long long> a,
    pair<signed long long, signed long long> b) {
  if (a.first > b.second || b.first > a.second) return {1, 0};
  return {max(a.first, b.first), min(a.second, b.second)};
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (signed long long i = 1; i <= n; i++) cin >> c[i];
  signed long long ans = 0;
  signed long long open;
  signed long long close = 0;
  for (signed long long i = 1; i <= n; i += 2) {
    open = close = 0;
    signed long long ok = 1;
    for (signed long long j = i + 1; j <= n; j++) {
      if (j & 1)
        open += c[j];
      else {
        signed long long d = open - close;
        pair<signed long long, signed long long> use =
            intersect({1, c[j]}, {d + ok, d + c[i]});
        if (use.second >= use.first) ans += use.second - use.first + 1;
        close += c[j];
        ok = max(ok, close - open);
        if (ok > c[i]) break;
      }
    }
  }
  cout << ans << "\n";
}

### Prompt

Create a solution in CPP for 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 signed long long MOD = 1e9 + 7;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
signed long long n;
signed long long c[1001];
pair<signed long long, signed long long> intersect(
    pair<signed long long, signed long long> a,
    pair<signed long long, signed long long> b) {
  if (a.first > b.second || b.first > a.second) return {1, 0};
  return {max(a.first, b.first), min(a.second, b.second)};
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  cin >> n;
  for (signed long long i = 1; i <= n; i++) cin >> c[i];
  signed long long ans = 0;
  signed long long open;
  signed long long close = 0;
  for (signed long long i = 1; i <= n; i += 2) {
    open = close = 0;
    signed long long ok = 1;
    for (signed long long j = i + 1; j <= n; j++) {
      if (j & 1)
        open += c[j];
      else {
        signed long long d = open - close;
        pair<signed long long, signed long long> use =
            intersect({1, c[j]}, {d + ok, d + c[i]});
        if (use.second >= use.first) ans += use.second - use.first + 1;
        close += c[j];
        ok = max(ok, close - open);
        if (ok > c[i]) break;
      }
    }
  }
  cout << ans << "\n";
}

```