Problem:
Score : 500 points

Problem Statement

You have a string A = A_1 A_2 ... A_n consisting of lowercase English letters.

You can choose any two indices i and j such that 1 \leq i \leq j \leq n and reverse substring A_i A_{i+1} ... A_j.

You can perform this operation at most once.

How many different strings can you obtain?

Constraints

1 \leq |A| \leq 200,000

A consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

A

Output

Print the number of different strings you can obtain by reversing any substring in A at most once.

Sample Input 1

aatt

Sample Output 1

5

You can obtain aatt (don't do anything), atat (reverse A[2..3]), atta (reverse A[2..4]), ttaa (reverse A[1..4]) and taat (reverse A[1..3]).

Sample Input 2

xxxxxxxxxx

Sample Output 2

1

Whatever substring you reverse, you'll always get xxxxxxxxxx.

Sample Input 3

abracadabra

Sample Output 3

44

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

#include<bits/stdc++.h>
#define rep(i,m) for(long long i=0; i<m; i++)
#define per(i,m) for(long long i=m-1; i>=0; i--)
#define FOR(i,n,m) for(long long i=n; i<m; i++)
#define ROF(i,n,m) for(long long i=m-1; i>=n; i--)
#define SORT(v,n) do{sort(v,v+n);reverse(v,v+n);}while(0)
#define all(x) (x).begin(),(x).end()
#define MP make_pair
#define MT make_tuple
#define EPS (1e-7)
#define INF (1e18)
#define PI (acos(-1))
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
using namespace std;
typedef long long ll;
const ll MOD = 1000000007;
typedef pair<int,int> P;
typedef pair<ll,ll> LP;

 
ll POW(ll x,ll n){
  x%=MOD;
  if(n==0)return 1;
  if(n%2==0)return POW(x*x,n/2)%MOD;
  return x%MOD*POW(x,n-1)%MOD;
}
 
ll POW2(ll x,ll n){
  if(n==0)return 1;
  if(n%2==0)return POW2(x*x,n/2);
  return x*POW2(x,n-1);
}
 
ll POW3(ll x,ll n,ll m){
  x%=m;
  if(n==0)return 1;
  if(n%2==0)return POW3(x*x,n/2,m)%m;
  return x*POW3(x,n-1,m)%m;
}
 
ll gcd(ll u, ll v) {
  ll r;
  while (0 != v) {
    r = u % v; u = v; v = r;
  }
  return u;
}
 
ll lcm(ll u, ll v) {
  return u/gcd(u,v)*v;
}
 
ll KAI(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI(m-1)%MOD;
}
 
ll KAI2(ll m)
{
  if(m<0) return 0;
  if(m==0) return 1;
  return m*KAI2(m-1);
}
 
ll extGCD(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll d = extGCD(b, a%b, y, x);
    y -= a / b * x;
    return d;
}
 
inline ll mod(ll a, ll m) {
    return (a % m + m) % m;
}
 
ll modinv(ll a) {
    ll x, y;
    extGCD(a, MOD, x, y);
    return mod(x, MOD);
}
 
ll COM(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI(m)%MOD*modinv(KAI(n)%MOD*KAI(m-n)%MOD)%MOD;
}
 
ll COM2(ll m,ll n)
{
  if(m<n) return 0;
  if(n<0) return 0;
  if(n==0) return 1;
  if(m==n) return 1;
  return KAI2(m)/KAI2(n)/KAI2(m-n);
}
 
ll DEC(ll x,ll m,ll n)//xのm進数でのx^nの位の値
{
  return x%POW2(m,n+1)/POW2(m,n);
}
 
ll keta(ll x,ll n)//xのn進数での桁数
{
  if(x==0)return 0;
  return keta(x/n,n)+1;
}
 
ll DIV(ll x,ll n)//x!のnで割り切れる回数
{
  if(x==0)return 0;
  return x/n+DIV(x/n,n);
}
 
ll ORD(ll x,ll n)//xのnで割り切れる回数
{
  if(x==0)return INF;
  if(x%n!=0)return 0;
  return 1+ORD(x/n,n);
}
 
ll SUP(ll x,ll n)//xのnで割れなくなるまで割ったときの余り
{
  if(x==0)return 0;
  if(x%n!=0)return x;
  return SUP(x/n,n);
}
 
ll SGS(ll x,ll y, ll m)//1+x+…+x^(y-1)をmで割った余り
{
  if(y==0)return 0;
  if(y%2==0){
    return (1+POW3(x,y/2,m))*SGS(x,y/2,m)%m;
  }
  return (1+x*SGS(x,y-1,m))%m;
}
 
ll SSGS(ll x,ll y,ll m)//Σ[k=1→y](1+x+…+x^(k-1))をmで割った余り
{
  if(y==0)return 0;
  if(y==1)return 1;
  if(y%2==0){
    return (SSGS(x,y/2,m)*(POW3(x,y/2,m)+1)%m+SGS(x,y/2,m)*y/2%m)%m;
  }
  return (SSGS(x,y-1,m)*x%m+y)%m;
}
 
void shuffle(ll array[], ll size) {
    for(ll i = 0; i < size; i++) {
        ll j = rand()%size;
        ll t = array[i];
        array[i] = array[j];
        array[j] = t;
    }
}

ll SQRT(ll n){
  ll ok,ng,mid;
  ng=n+1;
  if(303700500<ng)ng=303700500;
  ok=0;
  while(abs(ok-ng)>1){
    mid=(ok+ng)/2;
    if(mid*mid<=n){
      ok=mid;
    }
    else{
      ng=mid;
    }
  }
  return ok;
}
 
struct UnionFind
{
  vector<int> par;
  vector<int> sizes;
  UnionFind(int n) : par(n), sizes(n, 1) {
    rep(i,n) par[i] = i;
  }
  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    if (sizes[x] < sizes[y]) swap(x, y);
    par[y] = x;
    sizes[x] += sizes[y];
  }
  bool same(int x, int y) {
    return find(x) == find(y);
  }
  int size(int x) {
    return sizes[find(x)];
  }
};

map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}

bool is_prime(int64_t x) {
  if(x==1)return false;
  for(int64_t i = 2; i * i <= x; i++) {
    if(x % i == 0) return false;
  }
  return true;
}

struct edge{ll to, cost;};
struct graph{
  ll V;
  vector<vector<edge> > G;
  vector<ll> d;

  graph(ll n){
    init(n);
  }

  void init(ll n){
    V = n;
    G.resize(V);
    d.resize(V);
    rep(i,V){
      d[i] = INF;
    }
  }

  void add_edge(ll s, ll t, ll cost){
    edge e;
    e.to = t, e.cost = cost;
    G[s].push_back(e);
  }

  void dijkstra(ll s){
    rep(i,V){
      d[i] = INF;
    }
    d[s] = 0;
    priority_queue<LP,vector<LP>, greater<LP> > que;
    que.push(LP(0,s));
    while(!que.empty()){
      LP p = que.top(); que.pop();
      ll v = p.second;
      if(d[v]<p.first) continue;
      for(auto e : G[v]){
        if(d[e.to]>d[v]+e.cost){
          d[e.to] = d[v]+e.cost;
          que.push(LP(d[e.to],e.to));
        }
      }
    }
  }
};

ll d[300][300];

void warshall_floyd(ll n){  
  rep(i,n)rep(j,n)rep(k,n)d[j][k]=min(d[j][k],d[j][i]+d[i][k]);
}

struct bit{
  ll m;
  vector<ll> b;
  bit(ll i){
    m=i;
    b.resize(m+1);
  }
  ll num(ll i){
    return b[i];
  }
  ll sum(ll i){
    ll s=0;
    while(i>0){ 
      s+=b[i];
      i-=i&-i;
    }
    return s;
  }
  void add(ll i, ll x){
    while(i<=m){
      b[i]+=x;
      i+=i&-i;
    }
  }
};

int main(){
  ll a[30]={},ans=0,n;
  string s;
  cin >> s;
  n=s.size();
  rep(i,n){
    a[s[i]-'a']++;
  }
  ans=n*(n-1)/2+1;
  rep(i,26)ans-=a[i]*(a[i]-1)/2;
  printf("%lld",ans);
}