#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  {
    long long n;
    cin >> n;
    vector<long long> a(n + 1);
    for (long long i = 1; i <= n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) a[i] *= -1;
    }
    vector<long long> pref(n + 1);
    for (long long i = 1; i <= n; ++i) pref[i] = pref[i - 1] + a[i];
    long long ans = 0;
    for (long long l = 1; l <= n; l += 2) {
      long long cur = 0;
      long long mn = a[l] - 1;
      for (long long r = l + 1; r <= n; r += 2) {
        cur += a[r - 1];
        if (cur + a[r] >= 0) {
          long long tt = cur + a[r];
          if (tt <= mn) {
            ans++;
            long long rem = mn - tt;
            ans += min(abs(a[r] + 1), rem);
          }
        } else {
          if (cur > 0) {
            long long tmp = -cur;
            long long tt = cur + tmp;
            if (tt <= mn) {
              ans++;
              long long rem = mn - tt;
              ans += min(abs(tmp + 1), rem);
            }
          }
        }
        cur += a[r];
        mn = min(mn, cur);
        if (mn < 0) break;
        ;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  {
    long long n;
    cin >> n;
    vector<long long> a(n + 1);
    for (long long i = 1; i <= n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) a[i] *= -1;
    }
    vector<long long> pref(n + 1);
    for (long long i = 1; i <= n; ++i) pref[i] = pref[i - 1] + a[i];
    long long ans = 0;
    for (long long l = 1; l <= n; l += 2) {
      long long cur = 0;
      long long mn = a[l] - 1;
      for (long long r = l + 1; r <= n; r += 2) {
        cur += a[r - 1];
        if (cur + a[r] >= 0) {
          long long tt = cur + a[r];
          if (tt <= mn) {
            ans++;
            long long rem = mn - tt;
            ans += min(abs(a[r] + 1), rem);
          }
        } else {
          if (cur > 0) {
            long long tmp = -cur;
            long long tt = cur + tmp;
            if (tt <= mn) {
              ans++;
              long long rem = mn - tt;
              ans += min(abs(tmp + 1), rem);
            }
          }
        }
        cur += a[r];
        mn = min(mn, cur);
        if (mn < 0) break;
        ;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```