#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long INF = 1e18 + 7;
long long MOD = 998244353;
template <typename T>
T my_gcd(T a, T b) {
  return gcd(a, b);
}
template <typename T>
T my_min(T a, T b) {
  return min(a, b);
}
template <typename T>
T my_max(T a, T b) {
  return max(a, b);
}
template <typename T>
T my_and(T a, T b) {
  return (a & b);
}
template <typename T>
T my_xor(T a, T b) {
  return (a ^ b);
}
template <typename T>
T my_or(T a, T b) {
  return (a | b);
}
template <typename T>
T my_sum(T a, T b) {
  return (a + b);
}
template <typename T>
T my_sum_mod(T a, T b) {
  return (a + b) % MOD;
}
template <typename T>
T my_mod(T a, T mod) {
  return a % mod;
}
int solve() {
  long long n;
  while (!(cin >> n)) {
    return 1;
  }
  vector<long long> a(n + 1);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (int i = 0; i < n - 1; i += 2) {
    long long diff = 0, mn = INF;
    for (int j = i + 1; j < n; j += 2) {
      ans += max(0ll, min(1ll * a[i], a[j] - diff) + min(mn, diff - 1) + 1);
      mn = min(mn, diff - a[j]);
      diff += (-a[j] + a[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cout << std::fixed << std::setprecision(15);
  cin.tie(0);
  int TET = 1e9;
  for (int i = 1; i <= TET; i++) {
    if (solve()) {
      break;
    }
  }
}

### Prompt

Your task is to create 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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long INF = 1e18 + 7;
long long MOD = 998244353;
template <typename T>
T my_gcd(T a, T b) {
  return gcd(a, b);
}
template <typename T>
T my_min(T a, T b) {
  return min(a, b);
}
template <typename T>
T my_max(T a, T b) {
  return max(a, b);
}
template <typename T>
T my_and(T a, T b) {
  return (a & b);
}
template <typename T>
T my_xor(T a, T b) {
  return (a ^ b);
}
template <typename T>
T my_or(T a, T b) {
  return (a | b);
}
template <typename T>
T my_sum(T a, T b) {
  return (a + b);
}
template <typename T>
T my_sum_mod(T a, T b) {
  return (a + b) % MOD;
}
template <typename T>
T my_mod(T a, T mod) {
  return a % mod;
}
int solve() {
  long long n;
  while (!(cin >> n)) {
    return 1;
  }
  vector<long long> a(n + 1);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (int i = 0; i < n - 1; i += 2) {
    long long diff = 0, mn = INF;
    for (int j = i + 1; j < n; j += 2) {
      ans += max(0ll, min(1ll * a[i], a[j] - diff) + min(mn, diff - 1) + 1);
      mn = min(mn, diff - a[j]);
      diff += (-a[j] + a[j + 1]);
    }
  }
  cout << ans << endl;
  return 0;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cout << std::fixed << std::setprecision(15);
  cin.tie(0);
  int TET = 1e9;
  for (int i = 1; i <= TET; i++) {
    if (solve()) {
      break;
    }
  }
}

```