Problem:
Score : 100 points

Problem Statement

You are given a string S of length 3 consisting of a, b and c. Determine if S can be obtained by permuting abc.

Constraints

|S|=3

S consists of a, b and c.

Input

Input is given from Standard Input in the following format:

S

Output

If S can be obtained by permuting abc, print Yes; otherwise, print No.

Sample Input 1

bac

Sample Output 1

Yes

Swapping the first and second characters in bac results in abc.

Sample Input 2

bab

Sample Output 2

No

Sample Input 3

abc

Sample Output 3

Yes

Sample Input 4

aaa

Sample Output 4

No

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

#include <bits/stdc++.h>
using namespace std;
inline void ci(void){
  return;
}
template <typename First, typename... Rest>
void ci(First& first, Rest&... rest){
  cin >> first;
  ci(rest...);
  return;
}
inline void co(void){
  cout << "\n";
  return;
}
template <typename First, typename... Rest>
void co(First first, Rest... rest){
  cout << first << " ";
  co(rest...);
  return;
}
#define ll long long
#define ld long double
#define st string
#define dl double
#define wh while
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define tp tuple
#define ft first
#define sc second
#define ve vector
#define tos to_string
#define len length
#define rt return 0
#define br break
#define con continue
#define lcm(n, m) (n * m / gcd(n, m))
#define np next_permutation
#define repi(i, m, n) for(int i=(int)(m);i<(int)(n);i++)
#define rep(i, n) repi(i, 0, n)
#define repill(i, m, n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define repll(i, n) repill(i, 0, n)
#define revi(i, m, n) for(int i=(int)(m);i>(int)(n);i--)
#define rev(i, n) revi(i, 0, n)
#define revill(i, m, n) for(ll i=(ll)(m);i>(ll)(n);i--)
#define revll(i, n) revill(i, 0, n)
#define sz(v) ((int)(v).size()) 
#define each(i, n) for(auto&& i: n)
#define all(n) (n.begin(), n.end())
#define sp(n) cout << fixed << setprecision(n)
// sqrt ルート, int 2*10^9, gcd　最大公約数
// __builtin_popcount bitの個数

int main() {
  st S;
  ci(S);
  if(S == "abc" || S == "bca" || S == "cab" || S == "acb" || S == "bac" || S == "cba"){
    co("Yes");
  }
  else{
    co("No");
  }
}