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>
using namespace std;
const long long int maxn = 1e5 + 5;
const long long int inf = 1e9;
const long double PI = acos(-1);
void yes() { cout << "YES\n"; }
void no() { cout << "NO\n"; }
long long int pmod(long long int a, long long int mod = 1000000007) {
  return ((a) % mod + mod) % mod;
}
long long int cel(long long int x1, long long int y1) {
  if ((x1) <= 0) {
    return (x1) / (y1);
  }
  return (((x1) / (y1)) + (((x1) % (y1)) != 0));
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a / gcd(a, b)) * b;
}
long long bp(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long bpm(long long a, long long b, long long m = 1000000007) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res % m;
}
string vow = "aeiou";
const int dx4[] = {-1, 0, 1, 0};
const int dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy8[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int dxhorse[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dyhorse[] = {1, -1, 2, -2, 2, -2, 1, -1};
void solve() {
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  ;
  sort(arr, arr + n);
  long long int sum = 0;
  for (int i = 0; i < n; i++) sum += arr[i];
  if (sum % 2) {
    cout << "NO";
    return;
  }
  long long int sum1 = 0;
  for (int i = 0; i < n - 1; i++) sum1 += arr[i];
  if (arr[n - 1] > sum1) {
    cout << "NO";
    return;
  }
  cout << "YES";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  ;
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}
