Problem:
Score : 200 points

Problem Statement

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen -, and the other characters are digits from 0 through 9.

You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Constraints

1≤A,B≤5

|S|=A+B+1

S consists of - and digits from 0 through 9.

Input

Input is given from Standard Input in the following format:

A B
S

Output

Print Yes if S follows the postal code format in AtCoder Kingdom; print No otherwise.

Sample Input 1

3 4
269-6650

Sample Output 1

Yes

The (A+1)-th character of S is -, and the other characters are digits from 0 through 9, so it follows the format.

Sample Input 2

1 1
---

Sample Output 2

No

S contains unnecessary -s other than the (A+1)-th character, so it does not follow the format.

Sample Input 3

1 2
7444

Sample Output 3

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*---------------------------*\
____________Mari000____________         
________Mostafa  Shehab________    
\*---------------------------*/

#include <bits/stdc++.h>
using namespace std;
 
#define PI  3.14159265  // PI = acos(-1)
#define EPS (1e-10)
#define endl "\n"
typedef long long ll;
typedef long double ld;
//template <class Type>

ll power(ll x,ll y,ll p);
int Bit(ll x);
 
inline int D(){
	int t;
	scanf("%d",&t);
	return t;
}
inline ll llD(){
	ll t;
	scanf("%lld",&t);
	return t;
}

const int N=1e5 + 5;
int arr[N];
int n,m;
int a,b,c,d;
unordered_map<int,int> mp;
char ch[100];
bool f1=true;

int main(){
	scanf("%d%d",&a,&b);
	scanf("%s",ch);
	for(int i=0;ch[i];++i){
		if(i!=a)
			f1&=(ch[i]-'0'<10 && isdigit(ch[i]));
		else
			f1&=(ch[i] == '-');
			
	}
	puts(f1?"Yes" : "No");
		

}










ll power(ll x,ll y,ll p){ 
    ll res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or  
                // equal to p 
    while (y > 0){ 
        // If y is odd, multiply x with result 
        if (y & 1) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p;
    }
    return res; 
}


int Bit(ll x){
	if(!x)
		return 0;
	return 1+Bit(x>>1);
}