Problem:
Score : 100 points

Problem Statement

You are given a 4-character string S consisting of uppercase English letters.
Determine if S consists of exactly two kinds of characters which both appear twice in S.

Constraints

The length of S is 4.

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S consists of exactly two kinds of characters which both appear twice in S, print Yes; otherwise, print No.

Sample Input 1

ASSA

Sample Output 1

Yes

S consists of A and S which both appear twice in S.

Sample Input 2

STOP

Sample Output 2

No

Sample Input 3

FFEE

Sample Output 3

Yes

Sample Input 4

FREE

Sample Output 4

No

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

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

using namespace std;
#define int long long
#define ll long long
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define FORR(i,a,b)for(ll i=(a);i<=(b);i++)
#define repr(i,n) for(ll i=n;i>=0;i--)
#define P pair<ll, ll>
#define sz(x) (ll)x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<ll>
#define COUT(x) cout<<(x)<<endl
#define MA map<ll,ll>
#define SE set<ll>
#define PQ priority_queue<ll>
#define PQR priority_queue<ll,VE,greater<ll>>
#define COUT(x) cout<<(x)<<endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define EPS (1e-10)
#define pb push_back

const long long MOD = 1000000007;
//const long long MOD = 998244353;
const long long INF = 1LL << 60;
const double PI = acos(-1.0);


long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (b == 0) return a;
	unsigned r;
	while ((r = a % b)) {
		a = b;
		b = r;
	}
	return b;
}
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a * b / g;
}
bool prime(ll n) {
	for (ll i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}
map<ll, ll> prime_factor(ll n) {
	map<ll, ll> ret;
	for (ll i = 2; i * i <= n; i++) {
		while (n % i == 0) {
			ret[i]++;
			n /= i;
		}
	}
	if (n != 1) ret[n] = 1;
	return ret;
}
const int MAX = 2000005;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

using Graph = vector<VE>;
vector<pair<char, int>> RunLength(string s) {
	if (s.size() == 0)return {};
	vector<pair<char, int>>res(1, pair<char, int>(s[0], 0));
	for (char p : s) {
		if (res.back().first == p) {
			res.back().second++;
		}
		else {
			res.emplace_back(p, 1);
		}
	}
	return res;
}
const int mod = 1000000007;
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%mod + mod) % mod) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) {
		(x *= a.x) %= mod;
		return *this;
	}
	mint operator+(const mint a) const {
		mint res(*this);
		return res += a;
	}
	mint operator-(const mint a) const {
		mint res(*this);
		return res -= a;
	}
	mint operator*(const mint a) const {
		mint res(*this);
		return res *= a;
	}
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const {
		return pow(mod - 2);
	}
	mint& operator/=(const mint a) {
		return (*this) *= a.inv();
	}
	mint operator/(const mint a) const {
		mint res(*this);
		return res /= a;
	}
};
struct combination {
	vector<mint> fact, ifact;
	combination(int n) :fact(n + 1), ifact(n + 1) {
		//assert(n < mod);
		fact[0] = 1;
		for (int i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i;
		ifact[n] = fact[n].inv();
		for (int i = n; i >= 1; --i) ifact[i - 1] = ifact[i] * i;
	}
	mint operator()(int n, int k) {
		if (k < 0 || k > n) return 0;
		return fact[n] * ifact[k] * ifact[n - k];
	}
};
//comb(2000005);
// KETASUU
int GetDigit(int num) {
	return log10(num) + 1;
}
int cnt[30];
signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	string s; cin >> s;
	rep(i, 4)cnt[s[i] - 'A']++;
	int a = 0;
	bool ok = true;
	rep(i, 30) {
		if (cnt[i] > 0) {
			a++;
			if (cnt[i] != 2)ok = false;
		}
	}
	Yes(a == 2 && ok);
	return 0;
}