#include <bits/stdc++.h>
template <typename T>
bool chkmin(T &a, T b) {
  return (b < a) ? a = b, 1 : 0;
}
template <typename T>
bool chkmax(T &a, T b) {
  return (b > a) ? a = b, 1 : 0;
}
long long ksm(long long a, long long b) {
  if (b == 0) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 1000000007;
  if (b & 1) ns = ns * a % 1000000007;
  return ns;
}
using namespace std;
const int maxn = 1005;
int x[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i += 2) {
    long long sum = 0, lb = 0;
    for (int j = i + 1; j <= n; j++) {
      if (j % 2 == 0) {
        long long dif = -sum;
        long long l = -lb;
        l = std::max(l, 1ll);
        l = std::max(l, 1 + dif);
        long long r = x[i];
        r = std::min(r, x[j] + dif);
        if (r >= l) ans += r - l + 1;
      }
      if (j % 2)
        sum += x[j];
      else
        sum -= x[j];
      lb = std::min(lb, sum);
    }
  }
  cout << ans << endl;
  return (0 - 0);
}

### 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>
template <typename T>
bool chkmin(T &a, T b) {
  return (b < a) ? a = b, 1 : 0;
}
template <typename T>
bool chkmax(T &a, T b) {
  return (b > a) ? a = b, 1 : 0;
}
long long ksm(long long a, long long b) {
  if (b == 0) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 1000000007;
  if (b & 1) ns = ns * a % 1000000007;
  return ns;
}
using namespace std;
const int maxn = 1005;
int x[maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i += 2) {
    long long sum = 0, lb = 0;
    for (int j = i + 1; j <= n; j++) {
      if (j % 2 == 0) {
        long long dif = -sum;
        long long l = -lb;
        l = std::max(l, 1ll);
        l = std::max(l, 1 + dif);
        long long r = x[i];
        r = std::min(r, x[j] + dif);
        if (r >= l) ans += r - l + 1;
      }
      if (j % 2)
        sum += x[j];
      else
        sum -= x[j];
      lb = std::min(lb, sum);
    }
  }
  cout << ans << endl;
  return (0 - 0);
}

```