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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class _980_A_LinksAndPearls {

	public static void main(String[] args) throws IOException {
		String s = read(); int cnt1 = 0, cnt2 = 0; for(int i = 0; i<s.length(); i++) if(s.charAt(i) == 'o') cnt1++; else cnt2++;
		println(cnt1 == 0 || cnt2%cnt1 == 0 ? "YES" : "NO");
		exit();
	}

	final private static int BUFFER_SIZE = 1 << 16;
	private static DataInputStream din = new DataInputStream(System.in);
	private static byte[] buffer = new byte[BUFFER_SIZE];
	private static int bufferPointer = 0, bytesRead = 0;
	static PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

	public static String readLine() throws IOException {
		byte[] buf = new byte[64]; // line length
		int cnt = 0, c;
		while ((c = Read()) != -1) {
			if (c == '\n')
				break;
			buf[cnt++] = (byte) c;
		}
		return new String(buf, 0, cnt);
	}

	public static String read() throws IOException {
		byte[] ret = new byte[1024];
		int idx = 0;
		byte c = Read();
		while (c <= ' ') {
			c = Read();
		}
		do {
			ret[idx++] = c;
			c = Read();
		} while (c != -1 && c != ' ' && c != '\n' && c != '\r');
		return new String(ret, 0, idx);
	}

	public static int readInt() throws IOException {
		int ret = 0;
		byte c = Read();
		while (c <= ' ')
			c = Read();
		boolean neg = (c == '-');
		if (neg)
			c = Read();
		do {
			ret = ret * 10 + c - '0';
		} while ((c = Read()) >= '0' && c <= '9');

		if (neg)
			return -ret;
		return ret;
	}

	public static long readLong() throws IOException {
		long ret = 0;
		byte c = Read();
		while (c <= ' ')
			c = Read();
		boolean neg = (c == '-');
		if (neg)
			c = Read();
		do {
			ret = ret * 10 + c - '0';
		} while ((c = Read()) >= '0' && c <= '9');
		if (neg)
			return -ret;
		return ret;
	}

	public static double readDouble() throws IOException {
		double ret = 0, div = 1;
		byte c = Read();
		while (c <= ' ')
			c = Read();
		boolean neg = (c == '-');
		if (neg)
			c = Read();

		do {
			ret = ret * 10 + c - '0';
		} while ((c = Read()) >= '0' && c <= '9');

		if (c == '.') {
			while ((c = Read()) >= '0' && c <= '9') {
				ret += (c - '0') / (div *= 10);
			}
		}

		if (neg)
			return -ret;
		return ret;
	}

	private static void fillBuffer() throws IOException {
		bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
		if (bytesRead == -1)
			buffer[0] = -1;
	}

	private static byte Read() throws IOException {
		if (bufferPointer == bytesRead)
			fillBuffer();
		return buffer[bufferPointer++];
	}

	static void print(Object o) {
		pr.print(o);
	}

	static void println(Object o) {
		pr.println(o);
	}

	static void flush() {
		pr.flush();
	}

	static void println() {
		pr.println();
	}

	static void exit() throws IOException {
		din.close();
		pr.close();
		System.exit(0);
	}
}