Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about maximum values of the Main Uzhlyandian Function f, which is defined as follows:

<image>

In the above formula, 1 ≤ l < r ≤ n must hold, where n is the size of the Main Uzhlyandian Array a, and |x| means absolute value of x. But the heroes skipped their math lessons in school, so they asked you for help. Help them calculate the maximum value of f among all possible values of l and r for the given array a.

Input

The first line contains single integer n (2 ≤ n ≤ 105) — the size of the array a.

The second line contains n integers a1, a2, ..., an (-109 ≤ ai ≤ 109) — the array elements.

Output

Print the only integer — the maximum value of f.

Examples

Input

5
1 4 2 3 1


Output

3

Input

4
1 5 4 7


Output

6

Note

In the first sample case, the optimal value of f is reached on intervals [1, 2] and [2, 5].

In the second case maximal value of f is reachable only on the whole array.
#include <bits/stdc++.h>
enum { WHITE, GREY, BLACK };
using namespace std;
long long int inf = 1e18;
template <class T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T last_bit(T n) {
  return n & 1;
}
template <class T>
T big_mod(T n, T p, T m) {
  if (p == 0) return (T)1;
  T x = big_mod(n, p / 2, m);
  x = (x * x) % m;
  if (p & 1) x = (x * n) % m;
  return x;
}
template <class T>
T modInv(T a, T m) {
  T x, y;
  extgcd(a, m, x, y);
  x %= m;
  while (x < 0) {
    x += m;
  }
  return x;
}
template <class T>
T extgcd(T a, T b, T& x, T& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    int g = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return g;
  }
}
template <class T>
T multiplication(T n, T p, T m) {
  if (p == 0) return (T)0;
  T x = multiplication(n, p / 2, m);
  x = (x + x) % m;
  if (p & 1) x = (x + n) % m;
  return x;
}
template <class T>
T my_pow(T n, T p) {
  if (p == 0) return 1;
  T x = my_pow(n, p / 2);
  x = (x * x);
  if (p & 1) x = (x * n);
  return x;
}
template <class T>
double getdist(T a, T b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
template <class T>
T extract(string s, T ret) {
  stringstream ss(s);
  ss >> ret;
  return ret;
}
template <class T>
string tostring(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <class T>
T Mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
T MIN3(T a, T b, T c) {
  return min(a, min(b, c));
}
template <class T>
T MAX3(T a, T b, T c) {
  return max(a, max(b, c));
}
template <class T>
void print_vector(T& v) {
  int sz = v.size();
  if (sz) cout << v[0];
  for (int i = 1; i < sz; i++) cout << ' ' << v[i];
  cout << "\n";
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
bool isConsonant(char ch) {
  if (isalpha(ch) && !isVowel(ch)) return true;
  return false;
}
template <class R>
R Josephus(R n, R k) {
  R ans = 1;
  for (R i = 2; i <= n; i++) ans = (ans + k - 1) % i + 1;
  return ans;
}
template <class R>
R toitent_Phi2(R a) {
  R result = a;
  for (R i = 2; i * i <= a; i++) {
    if (a % i == 0) result = result - result / i;
    while (a % i == 0) a = a / i;
  }
  if (a > 1) result = result - result / a;
  return result;
}
template <typename T>
T Angle(T x1, T y1, T x2, T y2) {
  return atan(double(y1 - y2) / double(x1 - x2));
}
long long int ar[100009];
long long int dif1[100009], dif2[100009], dif[100009];
long long int kadane(long long int a[], long long int n) {
  long long int cur = 0;
  long long int mx = 0;
  for (long long int i = 1; i <= n; i++) {
    cur += a[i];
    if (cur < 0) cur = 0;
    if (cur > mx) mx = cur;
  }
  return mx;
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie();
  ;
  long long int n, i, j, k;
  scanf("%lld", &n);
  for (i = 1; i <= n; i++) scanf("%lld", &ar[i]);
  for (i = 1; i < n; i++) {
    dif[i] = abs(ar[i] - ar[i + 1]);
    dif1[i] = dif2[i] = dif[i];
  }
  for (i = 2; i < n; i = i + 2) dif1[i] = -dif1[i];
  for (i = 1; i < n; i = i + 2) dif2[i] = -dif2[i];
  printf("%lld\n", max(kadane(dif1, n - 1), kadane(dif2, n - 1)));
  return 0;
}
