#include <bits/stdc++.h>
using namespace std;
int read() {
  char ch = getchar();
  int x = 1, s = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return x * s;
}
long long a[100005];
int main() {
  int n = read();
  long long ans = 0;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int l = 1; l <= n; l += 2) {
    long long c = a[l], rem = 0, sum = 0;
    for (int r = l + 1; r <= n; r++) {
      if ((r - l) & 1) {
        sum += a[r];
        if (sum >= rem) {
          sum -= rem;
          rem = 0;
          if (c != a[l]) {
            ans++;
          }
          if (sum > c) {
            ans += c;
            break;
          } else {
            ans += sum;
            c -= sum;
            sum = 0;
          }
        } else {
          rem -= sum;
          sum = 0;
        }
      } else {
        rem += a[r];
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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;
int read() {
  char ch = getchar();
  int x = 1, s = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') x = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return x * s;
}
long long a[100005];
int main() {
  int n = read();
  long long ans = 0;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int l = 1; l <= n; l += 2) {
    long long c = a[l], rem = 0, sum = 0;
    for (int r = l + 1; r <= n; r++) {
      if ((r - l) & 1) {
        sum += a[r];
        if (sum >= rem) {
          sum -= rem;
          rem = 0;
          if (c != a[l]) {
            ans++;
          }
          if (sum > c) {
            ans += c;
            break;
          } else {
            ans += sum;
            c -= sum;
            sum = 0;
          }
        } else {
          rem -= sum;
          sum = 0;
        }
      } else {
        rem += a[r];
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```