#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chkmin(T &first, T second) {
  return second < first ? first = second, 1 : 0;
}
template <typename T>
bool chkmax(T &first, T second) {
  return first < second ? first = second, 1 : 0;
}
template <typename T>
void readint(T &first) {
  first = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) first = first * 10 + (c - '0');
  first *= f;
}
const int MAXN = 1005;
int n;
long long a[MAXN];
long long sum[MAXN][MAXN], minp[MAXN][MAXN];
int main() {
  readint(n);
  for (int i = 1; i <= n; ++i) readint(a[i]);
  if (n & 1) --n;
  long long res = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) {
      sum[i][j] = sum[i][j - 1] + a[j] * ((j & 1) ? 1 : -1);
      minp[i][j] = min(minp[i][j - 1], sum[i][j]);
    }
  for (int i = 1; i <= n; i += 2)
    for (int j = i + 1; j <= n; j += 2) {
      long long s = sum[i + 1][j - 1], m = -minp[i + 1][j - 1];
      long long l = m, r = a[i];
      chkmax(l, -s + 1);
      chkmin(r, a[j] - s);
      res += max(r - l + 1, 0ll);
    }
  cout << res << endl;
  return 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>
using namespace std;
template <typename T>
bool chkmin(T &first, T second) {
  return second < first ? first = second, 1 : 0;
}
template <typename T>
bool chkmax(T &first, T second) {
  return first < second ? first = second, 1 : 0;
}
template <typename T>
void readint(T &first) {
  first = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) first = first * 10 + (c - '0');
  first *= f;
}
const int MAXN = 1005;
int n;
long long a[MAXN];
long long sum[MAXN][MAXN], minp[MAXN][MAXN];
int main() {
  readint(n);
  for (int i = 1; i <= n; ++i) readint(a[i]);
  if (n & 1) --n;
  long long res = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j) {
      sum[i][j] = sum[i][j - 1] + a[j] * ((j & 1) ? 1 : -1);
      minp[i][j] = min(minp[i][j - 1], sum[i][j]);
    }
  for (int i = 1; i <= n; i += 2)
    for (int j = i + 1; j <= n; j += 2) {
      long long s = sum[i + 1][j - 1], m = -minp[i + 1][j - 1];
      long long l = m, r = a[i];
      chkmax(l, -s + 1);
      chkmin(r, a[j] - s);
      res += max(r - l + 1, 0ll);
    }
  cout << res << endl;
  return 0;
}

```