#include <bits/stdc++.h>
using namespace std;
mt19937 ran(time(0));
long long mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long Bpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans *= a;
      ans %= mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (long long i = 0; i < n; i += 2) {
    long long now = 0;
    long long mib = 0;
    for (long long j = i + 1; j < n; j++) {
      long long b = now + a[i] - a[j];
      now += (j % 2 ? -a[j] : a[j]);
      if (b > 0 && b > a[i] || b < 0 && -b > a[j] || j % 2 == 0) {
        mib = min(mib, now);
        continue;
      }
      long long l = a[i], r = a[j];
      if (b > 0)
        l -= b;
      else
        r += b;
      if (l > 0 && r > 0 && l >= -mib) ans += min(l + (mib ? mib + 1 : 0), r);
      mib = min(mib, now);
    }
  }
  cout << ans;
}

### Prompt

Please create a solution in CPP 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;
mt19937 ran(time(0));
long long mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long Bpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans *= a;
      ans %= mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (long long i = 0; i < n; i += 2) {
    long long now = 0;
    long long mib = 0;
    for (long long j = i + 1; j < n; j++) {
      long long b = now + a[i] - a[j];
      now += (j % 2 ? -a[j] : a[j]);
      if (b > 0 && b > a[i] || b < 0 && -b > a[j] || j % 2 == 0) {
        mib = min(mib, now);
        continue;
      }
      long long l = a[i], r = a[j];
      if (b > 0)
        l -= b;
      else
        r += b;
      if (l > 0 && r > 0 && l >= -mib) ans += min(l + (mib ? mib + 1 : 0), r);
      mib = min(mib, now);
    }
  }
  cout << ans;
}

```