#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool read(T &x) {
  int f = 1, c = getchar();
  x = 0;
  while (!isdigit(c)) {
    if (c == 45) f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x *= f, true;
}
template <typename T, typename... Args>
inline bool read(T &x, Args &...args) {
  return read(x) && read(args...);
}
namespace run {
const int maxn = 1e3 + 10;
long long a[maxn];
long long ans, now, mini;
int n;
inline int main() {
  read(n);
  for (register int i = 1; i <= n; ++i) read(a[i]);
  for (register int i = 1; i <= n; i += 2) {
    now = 0, mini = 0;
    for (register int j = i + 1; j <= n; ++j) {
      if (now < mini) mini = now;
      if (!(j & 1)) {
        long long lb = -mini, rb = now + lb;
        if (lb <= a[i] && rb <= a[j]) {
          ans += min(a[i] - lb + 1, a[j] - rb + 1);
          if (!lb && !rb) --ans;
        }
        now -= a[j];
      }
      if (j & 1) now += a[j];
    }
  }
  printf("%lld\n", ans);
  return 0;
}
}  // namespace run
int main() { return run ::main(); }

### Prompt

Develop a solution in Cpp 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>
inline bool read(T &x) {
  int f = 1, c = getchar();
  x = 0;
  while (!isdigit(c)) {
    if (c == 45) f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x *= f, true;
}
template <typename T, typename... Args>
inline bool read(T &x, Args &...args) {
  return read(x) && read(args...);
}
namespace run {
const int maxn = 1e3 + 10;
long long a[maxn];
long long ans, now, mini;
int n;
inline int main() {
  read(n);
  for (register int i = 1; i <= n; ++i) read(a[i]);
  for (register int i = 1; i <= n; i += 2) {
    now = 0, mini = 0;
    for (register int j = i + 1; j <= n; ++j) {
      if (now < mini) mini = now;
      if (!(j & 1)) {
        long long lb = -mini, rb = now + lb;
        if (lb <= a[i] && rb <= a[j]) {
          ans += min(a[i] - lb + 1, a[j] - rb + 1);
          if (!lb && !rb) --ans;
        }
        now -= a[j];
      }
      if (j & 1) now += a[j];
    }
  }
  printf("%lld\n", ans);
  return 0;
}
}  // namespace run
int main() { return run ::main(); }

```