#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e18;
const int N = 3e5 + 5;
int rand(int a, int b) { return a + rand() % (b - a + 1); }
int brute(vector<int>& c) {
  int n = c.size();
  string s;
  for (int i = 0; i < n; ++i) {
    int x = c[i];
    if (i % 2) {
      while (x--) {
        s.push_back(')');
      }
    } else {
      while (x--) {
        s.push_back('(');
      }
    }
  }
  int cnt = 0;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    int sum = 0;
    for (int j = i; j < n; ++j) {
      if (s[j] == '(') {
        sum++;
      } else {
        sum--;
      }
      if (sum < 0) {
        break;
      }
      cnt += (sum == 0);
    }
  }
  return cnt;
}
void solve() {
  int n;
  cin >> n;
  vector<int> c(n);
  for (auto& it : c) {
    cin >> it;
  }
  long long answer = 0;
  for (int i = 0; i < n; i += 2) {
    long long curr = c[i];
    long long mn = INF;
    for (int j = i + 1; j < n; ++j) {
      if (j % 2 == 0) {
        curr += c[j];
      } else {
        curr -= c[j];
        if (curr < 0 && -curr < c[j]) {
          long long R = mn + 1;
          long long L = 1;
          answer += max(0LL, min(R - L + 1, c[j] + curr));
        }
        if (curr > 0 && curr <= mn) {
          long long R = mn + 1;
          long long L = curr + 1;
          answer += max(0LL, min(R - L + 1, 1LL * c[j]));
        }
        if (curr == 0) {
          answer += min({mn + 1, 1ll * c[i], 1ll * c[j]});
        }
      }
      mn = min(mn, curr);
      if (curr < 0) {
        break;
      }
    }
  }
  cout << answer << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  int i = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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 INF = 1e18;
const int N = 3e5 + 5;
int rand(int a, int b) { return a + rand() % (b - a + 1); }
int brute(vector<int>& c) {
  int n = c.size();
  string s;
  for (int i = 0; i < n; ++i) {
    int x = c[i];
    if (i % 2) {
      while (x--) {
        s.push_back(')');
      }
    } else {
      while (x--) {
        s.push_back('(');
      }
    }
  }
  int cnt = 0;
  n = s.size();
  for (int i = 0; i < n; ++i) {
    int sum = 0;
    for (int j = i; j < n; ++j) {
      if (s[j] == '(') {
        sum++;
      } else {
        sum--;
      }
      if (sum < 0) {
        break;
      }
      cnt += (sum == 0);
    }
  }
  return cnt;
}
void solve() {
  int n;
  cin >> n;
  vector<int> c(n);
  for (auto& it : c) {
    cin >> it;
  }
  long long answer = 0;
  for (int i = 0; i < n; i += 2) {
    long long curr = c[i];
    long long mn = INF;
    for (int j = i + 1; j < n; ++j) {
      if (j % 2 == 0) {
        curr += c[j];
      } else {
        curr -= c[j];
        if (curr < 0 && -curr < c[j]) {
          long long R = mn + 1;
          long long L = 1;
          answer += max(0LL, min(R - L + 1, c[j] + curr));
        }
        if (curr > 0 && curr <= mn) {
          long long R = mn + 1;
          long long L = curr + 1;
          answer += max(0LL, min(R - L + 1, 1LL * c[j]));
        }
        if (curr == 0) {
          answer += min({mn + 1, 1ll * c[i], 1ll * c[j]});
        }
      }
      mn = min(mn, curr);
      if (curr < 0) {
        break;
      }
    }
  }
  cout << answer << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  int i = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```