#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
using ll = long long int;
using ld = long double;
template <typename T, typename V>
using P = pair<T, V>;
template <typename T>
using V = vector<T>;
constexpr ll MOD = 1e9 + 7;
constexpr ll INF = 0x3f3f3f3f;
constexpr ld PI = 3.1415926535897932384626;
constexpr ld EPS = 1e-9;
constexpr ll MAX = 4e6 + 5;
void solve() {
  ll n;
  cin >> n;
  V<ll> c(n + 2, 0);
  for (ll i = (ll)(1); i < (ll)(n + 1); i++) cin >> c[i];
  ll ans = 0;
  for (ll i = (ll)(1); i < (ll)(n + 1); i++)
    if (i % 2) {
      ll cur = 0, mn = INF;
      for (ll j = (ll)(i); j < (ll)(n + 1); j++) {
        if (j % 2)
          cur += c[j];
        else {
          cur -= c[j];
          if (cur == 0)
            ans += min({mn + 1, c[i], c[j]});
          else if (cur > 0 && cur <= mn)
            ans += max(0LL, min({mn + 1 - cur, c[i] - cur, c[j]}));
          else if (cur < 0 && -cur < c[j])
            ans += max(0LL, min({mn + 1, c[i], c[j] + cur}));
        }
        if (cur < 0) break;
        mn = min(mn, cur);
      }
    }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ll t = 1;
  for (ll i = (ll)(1); i < (ll)(t + 1); i++) solve();
}

### Prompt

Develop a solution in cpp 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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
using ll = long long int;
using ld = long double;
template <typename T, typename V>
using P = pair<T, V>;
template <typename T>
using V = vector<T>;
constexpr ll MOD = 1e9 + 7;
constexpr ll INF = 0x3f3f3f3f;
constexpr ld PI = 3.1415926535897932384626;
constexpr ld EPS = 1e-9;
constexpr ll MAX = 4e6 + 5;
void solve() {
  ll n;
  cin >> n;
  V<ll> c(n + 2, 0);
  for (ll i = (ll)(1); i < (ll)(n + 1); i++) cin >> c[i];
  ll ans = 0;
  for (ll i = (ll)(1); i < (ll)(n + 1); i++)
    if (i % 2) {
      ll cur = 0, mn = INF;
      for (ll j = (ll)(i); j < (ll)(n + 1); j++) {
        if (j % 2)
          cur += c[j];
        else {
          cur -= c[j];
          if (cur == 0)
            ans += min({mn + 1, c[i], c[j]});
          else if (cur > 0 && cur <= mn)
            ans += max(0LL, min({mn + 1 - cur, c[i] - cur, c[j]}));
          else if (cur < 0 && -cur < c[j])
            ans += max(0LL, min({mn + 1, c[i], c[j] + cur}));
        }
        if (cur < 0) break;
        mn = min(mn, cur);
      }
    }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ll t = 1;
  for (ll i = (ll)(1); i < (ll)(t + 1); i++) solve();
}

```