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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define REV(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define ts to_string
#define all(v) (v).begin(), (v).end()
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vs vector <string>
#define vvs vector <vector<string>>
#define pq  priority_queue<int>
using ll = long long;
/* ----------------よく使う数字や配列----------------- */
int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };
const int INF = 2000000000;
constexpr ll mod = 1e9 + 7;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */
/*---------------------template----------------------*/
/*----------------------end--------------------------*/
/* --------------------ライブラリ--------------------*/
bool isprime(ll a)//素数判定、primeならtrue,違うならfalse
{
    if (a < 2) return false;
    else if (a == 2) return true;
    else if (a % 2 == 0) return false;
    double m = sqrt(a);
    for (int i = 3; i <= m; i += 2)
    {
        if (a % i == 0)
        {
            return false;
        }
    }
    // 素数である
    return true;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while (n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
/* ----------------------end----------------------- */
/*---------------------臨時関数----------------------*/
/*----------------------end-------------------------*/
signed main(){
  int A,B;string S;cin>>A>>B>>S;
  bool b=true;
  rep(i,A+B+1){
    if(i!=A){
      if(S[i]-'0'>=10||S[i]-'0'<0){
        b=false;
        break;
      }
    }
    else {
      if(S[i]!='-'){
        b=false;
        break;
      }
    }
  }
  if(b)puts("Yes");
  else puts("No");
}