#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

using pi = pair<int,int>;

// (行き先, 容量, コスト, 逆辺)
struct edge{ int to,cap,cost,rev; };

int V; // TODO:initialize
const int MAX_V = 70; // TODO:initialize
const int INF = 191919; // TODO:initialize
vector<edge> G[MAX_V];
int h[MAX_V]; // ポテンシャル
int dist[MAX_V];
int prevv[MAX_V], preve[MAX_V]; // 直前の頂点と辺

void add_edge(int from, int to, int cap, int cost){
    G[from].pb({to,cap,cost,(int)G[to].size()});
    G[to].pb({from,0,-cost,(int)G[from].size()-1});
}

// sからtへの流量fの最小費用流(不可能なら-1)
int min_cost_flow(int s, int t, int f, bool neg = false){
    int res = 0;
    fill(h,h+V,0);
    while(f>0){
        priority_queue<pi,vector<pi>,greater<pi>> pq;
        fill(dist,dist+V,INF);
        dist[s]=0;
        if(neg)
        {
            // bellman-fordでhを更新
            neg = false;
            bool update;
            do{
                update = false;
                rep(v,V){
                    if(dist[v] == INF) continue;
                    rep(i,G[v].size()){
                        edge &e = G[v][i];
                        if(e.cap>0 && dist[e.to]>dist[v]+e.cost){
                            dist[e.to]=dist[v]+e.cost;
                            prevv[e.to] = v;
                            preve[e.to] = i;
                            update = true;
                        }
                    }
                }
            }while(update);
        }
        else
        {
            // dijkstraでhを更新
            pq.push(pi(0,s));
            while(!pq.empty()){
                pi p = pq.top();
                pq.pop();
                int v = p.se;
                if(p.fi>dist[v]) continue;
                rep(i,G[v].size()){
                    edge &e = G[v][i];
                    if(e.cap>0 && dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
                        dist[e.to] = dist[v]+e.cost+h[v]-h[e.to];
                        prevv[e.to] = v;
                        preve[e.to] = i;
                        pq.push(pi(dist[e.to],e.to));
                    }
                }
            }
        }

        // これ以上流せない
        if(dist[t]==INF) return -1;

        rep(v,V) h[v] += dist[v];

        // s-t間の最短路に沿って目一杯流す
        int d=f;
        for(int v=t; v!=s; v=prevv[v]) d = min(d,G[prevv[v]][preve[v]].cap);
        f -= d;
        res += d*h[t];

        for(int v=t; v!=s; v=prevv[v]){
            edge &e = G[prevv[v]][preve[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}

int main(){
    int n;
    string s,t;
    cin >>n >>s >>t;

    const int A = 26;
    vector<int> cs(A), ct(A);
    rep(i,n){
        ++cs[s[i]-'a'];
        ++ct[t[i]-'a'];
    }

    int S = 2*A, T = 2*A+1;
    V = 2*A+2;

    rep(i,A){
        add_edge(S,i,1,0);
        add_edge(A+i,T,1,0);
    }

    rep(i,A)rep(j,A){
        int d = abs(cs[i]-ct[j]);
        add_edge(i,A+j,1,d);
    }

    int ans = min_cost_flow(S,T,A);
    cout << ans/2 << endl;
    return 0;
}


### Prompt

In Cpp, your task is to solve the following problem:

Problem

Jennifer and Marian presented Carla with the string S.
However, Carla is not happy to receive the string S.
I wanted the string T.
The three decided to work together to change the string S to the string T.


Jennifer first sorts the letters in any order.

Marian then exchanges the lowercase letters of the two alphabets any number of times.
This operation exchanges all the same characters in the string, for example:


* aab-> Swap a and b-> bba
* aab-> Swap a and c-> ccb



Finally, Carla replaces one character with another and repeats until T.
Jennifer and Marian decided to try to reduce the number of Carla replacements.
Find the minimum number of replacements that Carla makes.

Constraints

* 1 ≤ n ≤ 105
* S and T contain only'a'~'z'
* | S | = | T | = n

Input


n
S
T


The length n of the string is given on the first line.
The character string S is given on the second line, and the character string T is given on the third line.

Output

Print the minimum number of Carla replacements on one line.

Examples

Input

3
abc
xyz


Output

0


Input

5
aaabb
xyxyz


Output

1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

using pi = pair<int,int>;

// (行き先, 容量, コスト, 逆辺)
struct edge{ int to,cap,cost,rev; };

int V; // TODO:initialize
const int MAX_V = 70; // TODO:initialize
const int INF = 191919; // TODO:initialize
vector<edge> G[MAX_V];
int h[MAX_V]; // ポテンシャル
int dist[MAX_V];
int prevv[MAX_V], preve[MAX_V]; // 直前の頂点と辺

void add_edge(int from, int to, int cap, int cost){
    G[from].pb({to,cap,cost,(int)G[to].size()});
    G[to].pb({from,0,-cost,(int)G[from].size()-1});
}

// sからtへの流量fの最小費用流(不可能なら-1)
int min_cost_flow(int s, int t, int f, bool neg = false){
    int res = 0;
    fill(h,h+V,0);
    while(f>0){
        priority_queue<pi,vector<pi>,greater<pi>> pq;
        fill(dist,dist+V,INF);
        dist[s]=0;
        if(neg)
        {
            // bellman-fordでhを更新
            neg = false;
            bool update;
            do{
                update = false;
                rep(v,V){
                    if(dist[v] == INF) continue;
                    rep(i,G[v].size()){
                        edge &e = G[v][i];
                        if(e.cap>0 && dist[e.to]>dist[v]+e.cost){
                            dist[e.to]=dist[v]+e.cost;
                            prevv[e.to] = v;
                            preve[e.to] = i;
                            update = true;
                        }
                    }
                }
            }while(update);
        }
        else
        {
            // dijkstraでhを更新
            pq.push(pi(0,s));
            while(!pq.empty()){
                pi p = pq.top();
                pq.pop();
                int v = p.se;
                if(p.fi>dist[v]) continue;
                rep(i,G[v].size()){
                    edge &e = G[v][i];
                    if(e.cap>0 && dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
                        dist[e.to] = dist[v]+e.cost+h[v]-h[e.to];
                        prevv[e.to] = v;
                        preve[e.to] = i;
                        pq.push(pi(dist[e.to],e.to));
                    }
                }
            }
        }

        // これ以上流せない
        if(dist[t]==INF) return -1;

        rep(v,V) h[v] += dist[v];

        // s-t間の最短路に沿って目一杯流す
        int d=f;
        for(int v=t; v!=s; v=prevv[v]) d = min(d,G[prevv[v]][preve[v]].cap);
        f -= d;
        res += d*h[t];

        for(int v=t; v!=s; v=prevv[v]){
            edge &e = G[prevv[v]][preve[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}

int main(){
    int n;
    string s,t;
    cin >>n >>s >>t;

    const int A = 26;
    vector<int> cs(A), ct(A);
    rep(i,n){
        ++cs[s[i]-'a'];
        ++ct[t[i]-'a'];
    }

    int S = 2*A, T = 2*A+1;
    V = 2*A+2;

    rep(i,A){
        add_edge(S,i,1,0);
        add_edge(A+i,T,1,0);
    }

    rep(i,A)rep(j,A){
        int d = abs(cs[i]-ct[j]);
        add_edge(i,A+j,1,d);
    }

    int ans = min_cost_flow(S,T,A);
    cout << ans/2 << endl;
    return 0;
}


```