Wilbur the pig is tinkering with arrays again. He has the array a1, a2, ..., an initially consisting of n zeros. At one step, he can choose any index i and either add 1 to all elements ai, ai + 1, ... , an or subtract 1 from all elements ai, ai + 1, ..., an. His goal is to end up with the array b1, b2, ..., bn. 

Of course, Wilbur wants to achieve this goal in the minimum number of steps and asks you to compute this value.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the length of the array ai. Initially ai = 0 for every position i, so this array is not given in the input.

The second line of the input contains n integers b1, b2, ..., bn ( - 109 ≤ bi ≤ 109).

Output

Print the minimum number of steps that Wilbur needs to make in order to achieve ai = bi for all i.

Examples

Input

5
1 2 3 4 5


Output

5

Input

4
1 2 2 1


Output

3

Note

In the first sample, Wilbur may successively choose indices 1, 2, 3, 4, and 5, and add 1 to corresponding suffixes.

In the second sample, Wilbur first chooses indices 1 and 2 and adds 1 to corresponding suffixes, then he chooses index 4 and subtract 1.
#include <bits/stdc++.h>
using namespace std;
using namespace chrono;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
void extendgcd(long long a, long long b, long long *v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long mminv(long long a, long long b) {
  long long arr[3];
  extendgcd(a, b, arr);
  return arr[0];
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
bool revsort(long long a, long long b) { return a > b; }
void swap(int &x, int &y) {
  int temp = x;
  x = y;
  y = temp;
}
long long combination(long long n, long long r, long long m, long long *fact,
                      long long *ifact) {
  long long val1 = fact[n];
  long long val2 = ifact[n - r];
  long long val3 = ifact[r];
  return (((val1 * val2) % m) * val3) % m;
}
void google(int t) { cout << "Case #" << t << ": "; }
vector<long long> sieve(int n) {
  int *arr = new int[n + 1]();
  vector<long long> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long mod_add(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long mod_mul(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long mod_sub(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long mod_div(long long a, long long b, long long m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mminvprime(b, m), m) + m) % m;
}
long long phin(long long n) {
  long long number = n;
  if (n % 2 == 0) {
    number /= 2;
    while (n % 2 == 0) n /= 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    if (n % i == 0) {
      while (n % i == 0) n /= i;
      number = (number / i * (i - 1));
    }
  }
  if (n > 1) number = (number / n * (n - 1));
  return number;
}
int isprime(int N) {
  if (N < 2 || (!(N & 1) && N != 2)) return 0;
  for (int i = 3; i * i <= N; i += 2) {
    if (!(N % i)) return 0;
  }
  return 1;
}
int bin_pow(int ind, int power) {
  if (power == 0) {
    return 1;
  }
  if (power & 1) {
    return ind * bin_pow(ind, power - 1) % 1000000007;
  } else {
    return bin_pow((ind * ind) % 1000000007, power / 2) % 1000000007;
  }
}
int fact(int x) {
  int ret = 1;
  for (int i = 2; i <= x; i++) {
    ret *= i;
    ret %= 1000000007;
  }
  return ret;
}
int ncr(int n, int r) {
  if (r > n) return 0;
  return (fact(n) *
          (bin_pow((fact(r) * fact(n - r)) % 1000000007, 1000000007 - 2))) %
         1000000007;
}
void solve() {
  int n;
  cin >> n;
  long long arr[200010], total = 0;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    if (i > 0)
      total += abs(arr[i] - arr[i - 1]);
    else
      total += abs(arr[i]);
  }
  cout << total << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  auto start1 = high_resolution_clock::now();
  solve();
  auto stop1 = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop1 - start1);
  return 0;
}
