#include <bits/stdc++.h>
using namespace std;
const long long MOD1 = 1000000000;
const long long INF = 1000000000000000;
long long vsum(vector<long long> &x) {
  long long sum = 0;
  for (auto itr1000 : x) sum += itr1000;
  return sum;
}
long long vsum(vector<long long> &x, long long MODULO) {
  long long sum = 0;
  for (auto itr1000 : x) {
    sum += itr1000;
    sum %= MODULO;
  };
  return sum;
}
long long binexpo(long long a, long long n, long long mod1000) {
  if (n <= 1) return a;
  if (n % 2) return ((a * binexpo(a, n - 1, mod1000)) % mod1000);
  long long res = (binexpo(a, n / 2, mod1000)) % mod1000;
  return (res * res) % mod1000;
}
long long modInverse(long long a, long long mod) {
  return binexpo(a, mod - 2, mod);
}
vector<long long> HAHA;
long long factorial(long long n) {
  if (!HAHA.size()) HAHA.push_back(1);
  for (long long itr1000 = HAHA.size(); itr1000 <= n; itr1000++)
    HAHA.push_back((itr1000 * HAHA[itr1000 - 1]) % MOD1);
  return HAHA[n];
}
long long nCr(long long n, long long r) {
  long long res = factorial(n) * modInverse(factorial(n - r), MOD1);
  res %= MOD1;
  res *= modInverse(factorial(r), MOD1);
  res %= MOD1;
  return res;
}
long long nPr(long long n, long long r) {
  long long res = factorial(n) * modInverse(factorial(n - r), MOD1);
  res %= MOD1;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long cnt = 0;
    long long mi = INF;
    for (long long j = i; j < n; j++) {
      if ((j - i) % 2) {
        ans += max(0ll, min(cnt, a[j]) - (cnt - min(a[i], mi)) + 1);
        if ((cnt - min(a[i], mi)) == 0) ans--;
        cnt -= a[j];
      } else {
        cnt += a[j];
      }
      mi = min(mi, cnt);
      if (mi < 0) break;
    }
    i++;
  }
  cout << ans << "\n";
}
bool cases = 0;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  if (cases) cin >> test;
  while (test--) solve();
}

### Prompt

Generate 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 MOD1 = 1000000000;
const long long INF = 1000000000000000;
long long vsum(vector<long long> &x) {
  long long sum = 0;
  for (auto itr1000 : x) sum += itr1000;
  return sum;
}
long long vsum(vector<long long> &x, long long MODULO) {
  long long sum = 0;
  for (auto itr1000 : x) {
    sum += itr1000;
    sum %= MODULO;
  };
  return sum;
}
long long binexpo(long long a, long long n, long long mod1000) {
  if (n <= 1) return a;
  if (n % 2) return ((a * binexpo(a, n - 1, mod1000)) % mod1000);
  long long res = (binexpo(a, n / 2, mod1000)) % mod1000;
  return (res * res) % mod1000;
}
long long modInverse(long long a, long long mod) {
  return binexpo(a, mod - 2, mod);
}
vector<long long> HAHA;
long long factorial(long long n) {
  if (!HAHA.size()) HAHA.push_back(1);
  for (long long itr1000 = HAHA.size(); itr1000 <= n; itr1000++)
    HAHA.push_back((itr1000 * HAHA[itr1000 - 1]) % MOD1);
  return HAHA[n];
}
long long nCr(long long n, long long r) {
  long long res = factorial(n) * modInverse(factorial(n - r), MOD1);
  res %= MOD1;
  res *= modInverse(factorial(r), MOD1);
  res %= MOD1;
  return res;
}
long long nPr(long long n, long long r) {
  long long res = factorial(n) * modInverse(factorial(n - r), MOD1);
  res %= MOD1;
  return res;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    long long cnt = 0;
    long long mi = INF;
    for (long long j = i; j < n; j++) {
      if ((j - i) % 2) {
        ans += max(0ll, min(cnt, a[j]) - (cnt - min(a[i], mi)) + 1);
        if ((cnt - min(a[i], mi)) == 0) ans--;
        cnt -= a[j];
      } else {
        cnt += a[j];
      }
      mi = min(mi, cnt);
      if (mi < 0) break;
    }
    i++;
  }
  cout << ans << "\n";
}
bool cases = 0;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  if (cases) cin >> test;
  while (test--) solve();
}

```