#include <bits/stdc++.h>
using namespace std;
long long n, a[1010];
long long res[1010 * 1010];
long long bf() {
  int m = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < a[i]; j++) res[m++] = ((i % 2) ? -1 : 1);
  }
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int cv = res[i];
    if (cv < 0) continue;
    for (int j = i + 1; j < m; j++) {
      cv += res[j];
      if (cv < 0) break;
      if (cv == 0) ans++;
    }
  }
  return ans;
}
long long solve() {
  long long ans = 0;
  for (long long i = 0; i < n; i += 2) {
    long long between = 0;
    long long minbetween = 0;
    for (long long j = i + 1; j < n; j += 2) {
      long long rem1 = a[i], rem2 = a[j];
      rem1 += minbetween;
      long long cb = between - minbetween;
      rem2 -= cb;
      if (rem1 >= 0 && rem2 >= 0) {
        ans += min(rem1, rem2);
        if (minbetween != 0) ans++;
      }
      between -= a[j];
      minbetween = min(minbetween, between);
      if (j + 1 < n) between += a[j + 1];
    }
  }
  return ans;
}
void gen(int c, int mv, int fc) {
  if (c < 0) {
    if (solve() != bf()) {
      cout << solve() << " " << bf() << endl;
      for (int i = 0; i <= fc; i++) cout << a[i] << " ";
      cout << endl;
    }
    return;
  }
  for (int i = 1; i <= mv; i++) {
    a[c] = i;
    gen(c - 1, mv, fc);
  }
}
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  printf("%lld\n", solve());
}

### 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;
long long n, a[1010];
long long res[1010 * 1010];
long long bf() {
  int m = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < a[i]; j++) res[m++] = ((i % 2) ? -1 : 1);
  }
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    int cv = res[i];
    if (cv < 0) continue;
    for (int j = i + 1; j < m; j++) {
      cv += res[j];
      if (cv < 0) break;
      if (cv == 0) ans++;
    }
  }
  return ans;
}
long long solve() {
  long long ans = 0;
  for (long long i = 0; i < n; i += 2) {
    long long between = 0;
    long long minbetween = 0;
    for (long long j = i + 1; j < n; j += 2) {
      long long rem1 = a[i], rem2 = a[j];
      rem1 += minbetween;
      long long cb = between - minbetween;
      rem2 -= cb;
      if (rem1 >= 0 && rem2 >= 0) {
        ans += min(rem1, rem2);
        if (minbetween != 0) ans++;
      }
      between -= a[j];
      minbetween = min(minbetween, between);
      if (j + 1 < n) between += a[j + 1];
    }
  }
  return ans;
}
void gen(int c, int mv, int fc) {
  if (c < 0) {
    if (solve() != bf()) {
      cout << solve() << " " << bf() << endl;
      for (int i = 0; i <= fc; i++) cout << a[i] << " ";
      cout << endl;
    }
    return;
  }
  for (int i = 1; i <= mv; i++) {
    a[c] = i;
    gen(c - 1, mv, fc);
  }
}
int main() {
  scanf("%lld", &n);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  printf("%lld\n", solve());
}

```