#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < 48 || c > 57; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= 48 && c <= 57; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  f ? x = -x : x;
}
template <typename T, typename... L>
inline void read(T &x, L &...l) {
  read(x), read(l...);
}
int n;
long long s[10005];
int main() {
  read(n);
  for (int i = 1; i <= n; i++)
    read(s[i]), s[i] = i & 1 ? s[i] : -s[i], s[i] += s[i - 1];
  long long rs = 0;
  for (int i = 1; i <= n; i += 2) {
    long long mn = 1e18;
    for (int j = i + 1; j <= n; j++) {
      long long qaq = s[j - 1] - s[i];
      if (!(j & 1) && max(1ll, -mn) <= s[i] - s[i - 1])
        rs += max(0ll, min(abs(s[i] - s[i - 1]), abs(s[j] - s[j - 1]) - qaq) -
                           max(1ll, -mn) + 1);
      mn = min(mn, s[j] - s[i]);
    }
  }
  return printf("%lld\n", rs), 0;
}

### Prompt

Your task is to create a Cpp solution to 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;
template <typename T>
inline void read(T &x) {
  x = 0;
  char c = getchar(), f = 0;
  for (; c < 48 || c > 57; c = getchar())
    if (!(c ^ 45)) f = 1;
  for (; c >= 48 && c <= 57; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  f ? x = -x : x;
}
template <typename T, typename... L>
inline void read(T &x, L &...l) {
  read(x), read(l...);
}
int n;
long long s[10005];
int main() {
  read(n);
  for (int i = 1; i <= n; i++)
    read(s[i]), s[i] = i & 1 ? s[i] : -s[i], s[i] += s[i - 1];
  long long rs = 0;
  for (int i = 1; i <= n; i += 2) {
    long long mn = 1e18;
    for (int j = i + 1; j <= n; j++) {
      long long qaq = s[j - 1] - s[i];
      if (!(j & 1) && max(1ll, -mn) <= s[i] - s[i - 1])
        rs += max(0ll, min(abs(s[i] - s[i - 1]), abs(s[j] - s[j - 1]) - qaq) -
                           max(1ll, -mn) + 1);
      mn = min(mn, s[j] - s[i]);
    }
  }
  return printf("%lld\n", rs), 0;
}

```