#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return x;
  return ~(x - 1);
}
int T, n;
long long s[1005], c[1005];
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    c[i] = read();
    if (i & 1)
      s[i] = s[i - 1] + (long long)c[i];
    else
      s[i] = s[i - 1] - (long long)c[i];
  }
  long long ans = 0ll;
  for (int i = 1; i <= n; i += 2) {
    long long ans1 = 0ll, minv = 1e18;
    for (int j = i + 1; j <= n; j++) {
      long long t = s[j - 1] - s[i];
      if (!(j & 1) && max(1ll, -minv) <= c[i] &&
          min(c[i], c[j] - t) >= max(1ll, -minv))
        ans += min(c[i], c[j] - t) - max(1ll, -minv) + 1;
      if (j & 1)
        ans1 += c[j];
      else
        ans1 -= c[j];
      minv = min(minv, ans1);
    }
  }
  cout << ans << "\n";
}

### 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;
inline long long read() {
  long long x = 0;
  bool flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + ch - '0';
    ch = getchar();
  }
  if (flag) return x;
  return ~(x - 1);
}
int T, n;
long long s[1005], c[1005];
signed main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    c[i] = read();
    if (i & 1)
      s[i] = s[i - 1] + (long long)c[i];
    else
      s[i] = s[i - 1] - (long long)c[i];
  }
  long long ans = 0ll;
  for (int i = 1; i <= n; i += 2) {
    long long ans1 = 0ll, minv = 1e18;
    for (int j = i + 1; j <= n; j++) {
      long long t = s[j - 1] - s[i];
      if (!(j & 1) && max(1ll, -minv) <= c[i] &&
          min(c[i], c[j] - t) >= max(1ll, -minv))
        ans += min(c[i], c[j] - t) - max(1ll, -minv) + 1;
      if (j & 1)
        ans1 += c[j];
      else
        ans1 -= c[j];
      minv = min(minv, ans1);
    }
  }
  cout << ans << "\n";
}

```