#include <bits/stdc++.h>
using namespace std;
const int N = 400010;
const int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double Pi = acos(-1.0);
int t, n;
long long a[N];
int dir[][2] = {{0, 0},  {0, 1},   {-1, 0}, {0, -1}, {1, 0},
                {-1, 1}, {-1, -1}, {1, -1}, {1, 1}};
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
inline int gcd(int x, int y) {
  if (!x) return y;
  return gcd(y % x, x);
}
inline int power(int x, int k, int p) {
  int res = 1;
  for (; k; k >>= 1, x = (long long)x * x % p)
    if (k & 1) res = (long long)res * x % p;
  return res;
}
long long k[N];
void solve() {
  n = read();
  long long r = 0, ans = 0;
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    if (i % 2) {
      r += a[i];
    } else {
      ans += min(r, a[i]);
      r = max((long long)0, r - a[i]);
      k[i] = 1;
      for (int j = i - 1; j >= 1; j -= 2) {
        if (a[j]) {
          int tmp = a[j];
          a[j] = max((long long)0, a[j] - a[i]);
          a[i] = max((long long)0, a[i] - tmp);
        }
        int las = j;
        if (a[i] == 0) {
          while (j >= 1 && a[j] == 0) j -= 2;
          for (int l = j; l < i; ++l) {
            ans += k[l];
          }
        }
        if (a[i] == 0) {
          for (int l = las; l < i; ++l) {
            k[l] = 0;
          }
          break;
        }
      }
      if (a[i] != 0) {
        for (int l = 1; l < i; ++l) {
          ans += k[l];
          k[l] = 0;
        }
        k[i] = 0;
      }
    }
  }
  cout << ans << endl;
  return;
}
int main() {
  solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

<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 int N = 400010;
const int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double Pi = acos(-1.0);
int t, n;
long long a[N];
int dir[][2] = {{0, 0},  {0, 1},   {-1, 0}, {0, -1}, {1, 0},
                {-1, 1}, {-1, -1}, {1, -1}, {1, 1}};
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
inline int gcd(int x, int y) {
  if (!x) return y;
  return gcd(y % x, x);
}
inline int power(int x, int k, int p) {
  int res = 1;
  for (; k; k >>= 1, x = (long long)x * x % p)
    if (k & 1) res = (long long)res * x % p;
  return res;
}
long long k[N];
void solve() {
  n = read();
  long long r = 0, ans = 0;
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    if (i % 2) {
      r += a[i];
    } else {
      ans += min(r, a[i]);
      r = max((long long)0, r - a[i]);
      k[i] = 1;
      for (int j = i - 1; j >= 1; j -= 2) {
        if (a[j]) {
          int tmp = a[j];
          a[j] = max((long long)0, a[j] - a[i]);
          a[i] = max((long long)0, a[i] - tmp);
        }
        int las = j;
        if (a[i] == 0) {
          while (j >= 1 && a[j] == 0) j -= 2;
          for (int l = j; l < i; ++l) {
            ans += k[l];
          }
        }
        if (a[i] == 0) {
          for (int l = las; l < i; ++l) {
            k[l] = 0;
          }
          break;
        }
      }
      if (a[i] != 0) {
        for (int l = 1; l < i; ++l) {
          ans += k[l];
          k[l] = 0;
        }
        k[i] = 0;
      }
    }
  }
  cout << ans << endl;
  return;
}
int main() {
  solve();
  return 0;
}

```