#include <bits/stdc++.h>
using namespace std;
const int mxn = (int)2e5 + 5;
void solve() {
  int n;
  cin >> n;
  int64_t ans = (int64_t)0;
  vector<int64_t> a(n);
  for (auto &x : a) cin >> x;
  vector<int> pos(n);
  vector<int64_t> stack;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      stack.push_back(a[i]);
    } else {
      int64_t cur = a[i];
      int ptr = -1;
      set<int> valid;
      bool block = true;
      for (int j = (int)stack.size() - 1; j >= 0; j -= 2) {
        int64_t take = min(cur, stack[j]);
        cur -= take;
        ans += take;
        if (take > (int64_t)0) {
          valid.insert(j);
        }
        if (stack[j] > take) {
          stack[j] -= take;
          ptr = j;
          pos[i] = j;
          break;
        } else {
          stack[j] = 0;
        }
        if (cur == 0) {
          pos[i] = j;
          block = false;
          break;
        }
      }
      if (block == false) {
        for (int j = (int)stack.size() - 1; j >= 0; j -= 2) {
          if (stack[j] > 0) {
            ptr = j;
            break;
          }
        }
      }
      int64_t cnt = (int64_t)0;
      for (int x : valid) {
        for (int j = x - 1; j > ptr;) {
          if (stack[j] > 0) {
            break;
          }
          cnt++;
          if (j - 1 >= 0 && stack[j - 1] > 0) break;
          j = pos[j] - 1;
        }
        ptr = x;
        if (0) cerr << x << " " << cnt << "\n";
      }
      ans += cnt;
      if (0)
        cerr << i << " " << ptr << " " << ans << " " << cnt << " " << cur << " "
             << pos[i] << "\n";
      stack.push_back(cur);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

<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 int mxn = (int)2e5 + 5;
void solve() {
  int n;
  cin >> n;
  int64_t ans = (int64_t)0;
  vector<int64_t> a(n);
  for (auto &x : a) cin >> x;
  vector<int> pos(n);
  vector<int64_t> stack;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      stack.push_back(a[i]);
    } else {
      int64_t cur = a[i];
      int ptr = -1;
      set<int> valid;
      bool block = true;
      for (int j = (int)stack.size() - 1; j >= 0; j -= 2) {
        int64_t take = min(cur, stack[j]);
        cur -= take;
        ans += take;
        if (take > (int64_t)0) {
          valid.insert(j);
        }
        if (stack[j] > take) {
          stack[j] -= take;
          ptr = j;
          pos[i] = j;
          break;
        } else {
          stack[j] = 0;
        }
        if (cur == 0) {
          pos[i] = j;
          block = false;
          break;
        }
      }
      if (block == false) {
        for (int j = (int)stack.size() - 1; j >= 0; j -= 2) {
          if (stack[j] > 0) {
            ptr = j;
            break;
          }
        }
      }
      int64_t cnt = (int64_t)0;
      for (int x : valid) {
        for (int j = x - 1; j > ptr;) {
          if (stack[j] > 0) {
            break;
          }
          cnt++;
          if (j - 1 >= 0 && stack[j - 1] > 0) break;
          j = pos[j] - 1;
        }
        ptr = x;
        if (0) cerr << x << " " << cnt << "\n";
      }
      ans += cnt;
      if (0)
        cerr << i << " " << ptr << " " << ans << " " << cnt << " " << cur << " "
             << pos[i] << "\n";
      stack.push_back(cur);
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```