#include <bits/stdc++.h>
using namespace std;
const long long inf = 1e9 + 7;
const long long linf = 1e18 + 1;
template <typename T>
using V = vector<T>;
template <typename T = long long>
T nxt() {
  T x;
  cin >> x;
  return x;
}
template <typename T = long long>
void read(vector<T>& x) {
  for (T& i : x) cin >> i;
}
template <typename T = long long>
void print(vector<T>& x) {
  for (T v : x) {
    cout << v << " ";
  }
}
template <typename T = long long>
void printset(set<T>& x) {
  for (auto v : x) cout << v << " ";
  cout << endl;
}
void print(map<char, long long>& cnt) {
  for (auto [ch, count] : cnt) {
    for (long long i = 0; i < count; ++i) {
      cout << ch;
    }
  }
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long TT = 1;
  while (TT--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    read(a);
    long long ans = 0;
    for (long long l = 0; l < n; l += 2) {
      long long fp = a[l];
      long long lp = 0;
      for (long long r = l + 1; r < n; ++r) {
        if (r % 2 == 0) {
          lp += a[r];
          continue;
        }
        long long v = a[r];
        long long d = min(v, lp);
        v -= d;
        lp -= d;
        if (v == 0) {
          if (lp == 0) {
            ++ans;
          }
          continue;
        }
        if (fp != a[l]) {
          ++ans;
        }
        ans += min(fp, v);
        fp -= v;
        if (fp < 0) {
          break;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
const long long inf = 1e9 + 7;
const long long linf = 1e18 + 1;
template <typename T>
using V = vector<T>;
template <typename T = long long>
T nxt() {
  T x;
  cin >> x;
  return x;
}
template <typename T = long long>
void read(vector<T>& x) {
  for (T& i : x) cin >> i;
}
template <typename T = long long>
void print(vector<T>& x) {
  for (T v : x) {
    cout << v << " ";
  }
}
template <typename T = long long>
void printset(set<T>& x) {
  for (auto v : x) cout << v << " ";
  cout << endl;
}
void print(map<char, long long>& cnt) {
  for (auto [ch, count] : cnt) {
    for (long long i = 0; i < count; ++i) {
      cout << ch;
    }
  }
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long TT = 1;
  while (TT--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    read(a);
    long long ans = 0;
    for (long long l = 0; l < n; l += 2) {
      long long fp = a[l];
      long long lp = 0;
      for (long long r = l + 1; r < n; ++r) {
        if (r % 2 == 0) {
          lp += a[r];
          continue;
        }
        long long v = a[r];
        long long d = min(v, lp);
        v -= d;
        lp -= d;
        if (v == 0) {
          if (lp == 0) {
            ++ans;
          }
          continue;
        }
        if (fp != a[l]) {
          ++ans;
        }
        ans += min(fp, v);
        fp -= v;
        if (fp < 0) {
          break;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```