You are given an array a_1, a_2, …, a_n.

In one operation you can choose two elements a_i and a_j (i ≠ j) and decrease each of them by one.

You need to check whether it is possible to make all the elements equal to zero or not.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5) — the size of the array.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the array.

Output

Print "YES" if it is possible to make all elements zero, otherwise print "NO".

Examples

Input


4
1 1 2 2


Output


YES

Input


6
1 2 3 4 5 6


Output


NO

Note

In the first example, you can make all elements equal to zero in 3 operations: 

  * Decrease a_1 and a_2, 
  * Decrease a_3 and a_4, 
  * Decrease a_3 and a_4 



In the second example, one can show that it is impossible to make all elements equal to zero.
#include <bits/stdc++.h>
const long long int N = 1e9 + 7;
const long long int mod = 1e9 + 7;
using namespace std;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
bool prime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool isPowerOfTwo(long long int n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
void SieveOfEratosthenes(long long int n) {
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (prime[p]) cout << p << " ";
}
long long int modularinv(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularinv((x * x) % mod, n / 2);
  else
    return (x * modularinv((x * x) % mod, (n - 1) / 2)) % mod;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
map<long long int, long long int> mpp;
void primeFactors(long long int n) {
  while (n % 2 == 0) {
    mpp[2]++;
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      mpp[i]++;
      n = n / i;
    }
  }
  if (n > 2) mpp[n]++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long int t, n, i, m, k, q, q1, l, j, d, r;
  t = 1;
  while (t--) {
    cin >> n;
    vector<long long int> v(n);
    long long int sum = 0;
    for (i = 0; i < n; i++) {
      cin >> v[i];
      sum += v[i];
    }
    sort(v.begin(), v.end());
    long long int d = sum;
    long long int flag = 0;
    long long int count = 0;
    for (i = 0; i < n - 1; i++) {
      count += v[i];
      if (v[n - 1] <= count) flag = 1;
    }
    if (d % 2 == 0) {
      if (flag == 1)
        cout << "YES" << '\n';
      else
        cout << "NO" << '\n';
    } else
      cout << "NO" << '\n';
  }
}
