Problem: A necklace can be described as a string of links ('-') and pearls ('o'), with the last link or pearl connected to the first one.

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You can remove a link or a pearl and insert it between two other existing links or pearls (or between a link and a pearl) on the necklace. This process can be repeated as many times as you like, but you can't throw away any parts.

Can you make the number of links between every two adjacent pearls equal? Two pearls are considered to be adjacent if there is no other pearl between them.

Note that the final necklace should remain as one circular part of the same length as the initial necklace.

Input

The only line of input contains a string s (3 ≤ |s| ≤ 100), representing the necklace, where a dash '-' represents a link and the lowercase English letter 'o' represents a pearl.

Output

Print "YES" if the links and pearls can be rejoined such that the number of links between adjacent pearls is equal. Otherwise print "NO".

You can print each letter in any case (upper or lower).

Examples

Input

<span class="tex-font-style-tt">-o-o--</span>

Output

YES

Input

<span class="tex-font-style-tt">-o---</span>


Output

YES

Input

<span class="tex-font-style-tt">-o---o-</span>


Output

NO

Input

ooo


Output

YES

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1000000007;
long long int inf = 1e15;
void scan(int &x);
long long int powermod(long long int _a, long long int _b, long long int _m) {
  long long int _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a) % _m;
    _b /= 2;
    _a = (_a * _a) % _m;
  }
  return _r;
}
long long int string_to_number(string s) {
  long long int x = 0;
  stringstream convert(s);
  convert >> x;
  return x;
}
long long int add(long long int a, long long int b) {
  long long int x = (a + b) % MOD;
  return x;
}
long long int mul(long long int a, long long int b) {
  long long int x = (a * b) % MOD;
  return x;
}
long long int sub(long long int a, long long int b) {
  long long int x = (a - b + MOD) % MOD;
  return x;
}
long long int divi(long long int a, long long int b) {
  long long int x = a;
  long long int y = powermod(b, MOD - 2, MOD);
  long long int res = (x * y) % MOD;
  return res;
}
int min(int a, int b) { return a < b ? a : b; }
int n, m;
bool bfs(string s1, string s2) {
  vector<vector<bool>> vis(n, vector<bool>(m));
  vis[0][0] = true;
  vector<pair<int, int>> q = {{0, 0}};
  for (int i = 0; i < q.size(); i++) {
    int x = q[i].first, y = q[i].second;
    int x2 = x, y2 = y;
    if (s1[x] == '>')
      y2++;
    else
      y2--;
    if (y2 >= 0 && y2 < m && !vis[x2][y2]) {
      vis[x2][y2] = true;
      q.push_back({x2, y2});
    }
    int x3 = x, y3 = y;
    if (s2[y] == 'v')
      x3++;
    else
      x3--;
    if (x3 >= 0 && x3 < n && !vis[x3][y3]) {
      vis[x3][y3] = true;
      q.push_back({x3, y3});
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (!vis[i][j]) return false;
  return true;
}
int main() {
  string s;
  cin >> s;
  int cnt_l = 0, cnt_p = 0;
  for (int i = 0; i < s.size(); i++) {
    if (s[i] == 'o')
      cnt_p++;
    else
      cnt_l++;
  }
  if (cnt_p == 0)
    cout << "YES";
  else if (cnt_l % cnt_p == 0)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}
void scan(int &x) {
  register int c = getchar();
  x = 0;
  int neg = 0;
  for (; ((c < 48 || c > 57) && c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}