#include <bits/stdc++.h>
using namespace std;
int n, a[100000], b[100000];
vector<int> v[2];
bool isprime(int k) {
  for (int i = 2; i <= sqrt(k); i++) {
    if (!(k % i)) {
      return false;
    }
  }
  return true;
}
int bm(int a, int b, int mod) {
  if (b == 0) {
    return 1;
  }
  int t = bm(a, b >> 1, mod);
  t = t * t % mod;
  if (b & 1) {
    return t * a % mod;
  }
  return t;
}
void eat() {
  long long c = 0;
  cin >> n;
  vector<pair<int, int>> v = {{0, 0}};
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 1) {
      while (!v.empty() && a[i] > 0) {
        if (a[i] < v.back().first) {
          v.back().first -= a[i];
          c = c + a[i] - 1 + (v.back().second * (v.back().second + 1)) / 2;
          v.back().second = 1;
          a[i] = 0;
        } else {
          c = c + v.back().first - 1 +
              (v.back().second * (v.back().second + 1)) / 2;
          a[i] -= v.back().first;
          v.pop_back();
          if (v.empty()) {
            v.push_back({0, 0});
            if (a[i] == 0) {
              v.back().second += 1;
            } else {
              c += 1;
            }
            break;
          }
          v.back().second += 1;
        }
      }
      if (v.empty()) {
        v.push_back({0, 0});
      }
    } else {
      v.push_back({a[i], 0});
    }
  }
  while (!v.empty()) {
    c = c + (v.back().second * (v.back().second + 1)) / 2;
    v.pop_back();
  }
  cout << c << '\n';
}
int main() {
  long long t = 1, ti;
  for (ti = 0; ti < t; ti++) {
    eat();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int n, a[100000], b[100000];
vector<int> v[2];
bool isprime(int k) {
  for (int i = 2; i <= sqrt(k); i++) {
    if (!(k % i)) {
      return false;
    }
  }
  return true;
}
int bm(int a, int b, int mod) {
  if (b == 0) {
    return 1;
  }
  int t = bm(a, b >> 1, mod);
  t = t * t % mod;
  if (b & 1) {
    return t * a % mod;
  }
  return t;
}
void eat() {
  long long c = 0;
  cin >> n;
  vector<pair<int, int>> v = {{0, 0}};
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i % 2 == 1) {
      while (!v.empty() && a[i] > 0) {
        if (a[i] < v.back().first) {
          v.back().first -= a[i];
          c = c + a[i] - 1 + (v.back().second * (v.back().second + 1)) / 2;
          v.back().second = 1;
          a[i] = 0;
        } else {
          c = c + v.back().first - 1 +
              (v.back().second * (v.back().second + 1)) / 2;
          a[i] -= v.back().first;
          v.pop_back();
          if (v.empty()) {
            v.push_back({0, 0});
            if (a[i] == 0) {
              v.back().second += 1;
            } else {
              c += 1;
            }
            break;
          }
          v.back().second += 1;
        }
      }
      if (v.empty()) {
        v.push_back({0, 0});
      }
    } else {
      v.push_back({a[i], 0});
    }
  }
  while (!v.empty()) {
    c = c + (v.back().second * (v.back().second + 1)) / 2;
    v.pop_back();
  }
  cout << c << '\n';
}
int main() {
  long long t = 1, ti;
  for (ti = 0; ti < t; ti++) {
    eat();
  }
  return 0;
}

```