1556_C. Compressed Bracket Sequence_160
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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.
#include <bits/stdc++.h>
using namespace std;
signed main() {
  setlocale(LC_ALL, "rus");
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  vector<long long> dp(n);
  stack<pair<int, int>> st;
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (i & 1) {
      while (!st.empty() && st.top().first < a[i]) {
        ans += st.top().first;
        if (st.top().second > 0) {
          ans += dp[st.top().second - 1];
        }
        a[i] -= st.top().first;
        st.pop();
      }
      if (!st.empty()) {
        ans += a[i];
        dp[i] = 1;
        if (a[i] == st.top().first) {
          if (st.top().second > 0) {
            ans += dp[st.top().second - 1];
            dp[i] += dp[st.top().second - 1];
          }
          st.pop();
        } else {
          st.top().first -= a[i];
        }
        a[i] = 0;
      }
    } else {
      st.emplace(a[i], i);
    }
  }
  cout << ans;
}
