#include <bits/stdc++.h>
using namespace std;
const long long N = 1e3 + 5;
const long long oo = 1e18 + 7, mod = 1e9 + 7;
long long n, a[N], pref[N], mn[N][N];
void process() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    if (i & 1)
      pref[i] = (pref[i - 1] + a[i]);
    else
      pref[i] = (pref[i - 1] - a[i]);
  }
  for (long long i = 1; i <= n; i++) {
    mn[i][i - 1] = oo;
    mn[i][i] = pref[i] - pref[i - 1];
    for (long long j = i + 1; j <= n; j++) {
      mn[i][j] = min(mn[i][j - 1], pref[j] - pref[i - 1]);
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = i + 1; j <= n; j++) {
      long long temp = 0;
      if (j > (i + 1)) temp = pref[j - 1] - pref[i];
      if (!(i & 1) || j & 1) continue;
      long long diff = -temp, mn_odd = max(0LL, -mn[i + 1][j - 1]),
                mx_odd = min(a[i], a[j] + diff);
      mn_odd = max(mn_odd, diff + 1);
      mn_odd = max(mn_odd, 1LL);
      ans += max(0LL, (mx_odd - mn_odd + 1));
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  process();
}

### Prompt

Please provide a CPP coded solution 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;
const long long N = 1e3 + 5;
const long long oo = 1e18 + 7, mod = 1e9 + 7;
long long n, a[N], pref[N], mn[N][N];
void process() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    if (i & 1)
      pref[i] = (pref[i - 1] + a[i]);
    else
      pref[i] = (pref[i - 1] - a[i]);
  }
  for (long long i = 1; i <= n; i++) {
    mn[i][i - 1] = oo;
    mn[i][i] = pref[i] - pref[i - 1];
    for (long long j = i + 1; j <= n; j++) {
      mn[i][j] = min(mn[i][j - 1], pref[j] - pref[i - 1]);
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = i + 1; j <= n; j++) {
      long long temp = 0;
      if (j > (i + 1)) temp = pref[j - 1] - pref[i];
      if (!(i & 1) || j & 1) continue;
      long long diff = -temp, mn_odd = max(0LL, -mn[i + 1][j - 1]),
                mx_odd = min(a[i], a[j] + diff);
      mn_odd = max(mn_odd, diff + 1);
      mn_odd = max(mn_odd, 1LL);
      ans += max(0LL, (mx_odd - mn_odd + 1));
    }
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0);
  process();
}

```