#include <bits/stdc++.h>
#include "dna.h"
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef pair<ll,ll> pll;
typedef tuple<int,int,int> ti;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<ll> vll;
typedef pair<ld,ld> pld;
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front
#define ff first
#define ss second
#define MOD (int)(1e8)
#define INF (ll) (1e18)
#define all(v) (v).begin(),(v).end()
ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;}
ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);}
ld pointdist(ld x, ld y, ld point) { return ((x-point)*(y-point))/abs(x-y); }
//ld dist(ld x, ld y, ld a, ld b){ return sqrt((x-a)*(x-a) + (y-b)*(y-b)); }
const int nx[8] = {0, 0, 1, -1,1,1,-1,-1}, ny[8] = {1, -1, 0, 0,1,-1,1,-1}; //East, West, South, North+
////////////******SOLUTION******\\\\\\\\\\\
const int MAX_N = 1e5 + 4;
string a, b;
int n;
int Apref[MAX_N][3];
int Bpref[MAX_N][3];
int cross[MAX_N][3][3];
int error[MAX_N];
char ar[3] = {'A','C','T'};
void build()
{
for(int k = 0; k < 3; k ++)
{
Apref[0][k] = Bpref[0][k] = 0;
for(int i = 0; i <n; i ++)
{
Apref[i+1][k] = Apref[i][k] + (a[i] == ar[k]);
Bpref[i+1][k] = Bpref[i][k] + (b[i] == ar[k]);
}
}
for(int j = 0; j < 3; j ++)
{
for(int k = 0; k < 3; k ++)
{
if(j == k) continue;
for(int i = 0; i <n; i ++)
cross[i+1][j][k] = cross[i][j][k] + (a[i] == ar[j] && b[i] == ar[k]);
}
}
for(int i = 0; i <n; i ++) error[i+1] = error[i] + (a[i] != b[i]);
}
void init(string ab, string bb)
{
a = ab; b = bb;
n = a.length();
build();
}
int get_distance(int l, int r)
{
for(int k = 0; k < 2; k ++)
if(Apref[r+1][k] - Apref[l][k] != Bpref[r+1][k] - Bpref[l][k])
return -1;
int faux = error[r+1] - error[l];
int dv = 0;
for(int j = 0; j <2; j ++)
{
for(int k = j+1; k <3; k ++)
{
int bg= min(cross[r+1][j][k] - cross[l][j][k], cross[r+1][k][j] - cross[l][k][j]);
bg *= 2;
dv += bg;
//cout << ar[j] << ' ' << ar[k] << ' ' << bg << '\n';
}
}
int cost = dv/2 + ((faux-dv)*2)/3 + ((faux-dv)%3 != 0);
return cost;
}
/*int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int p, q;
cin >> p >> q;
string x, y;
cin >> x >> y;
init(x,y);
while(q--)
{
int u, v;
cin >> u >> v;
cout << get_distance(u,v) << '\n' ;
}
/*for(int i = 1; i <=n; i ++)
{
for(int j = 0; j < 3; j ++)
{
for(int k = 0; k <3;k ++)
{
if(j == k) continue;
cout << ar[j] << '_' << ar[k] << ":" << cross[i][j][k] << ' ';
}
}
cout << '\n';
}
}
/*
6 3
ATACAT
ACTATA"
1 3
4 5
3 5
*/
/*
Identify problem diagram: Brute force, Greedy, Dynamic Programming, Divide and Conquer
Reformulate the problem into something more theoretical
!!!!! IMPLICIT GRAPH ??????
!!!!! PAY ATTENTION TO THE CONSTRAINTS: DP nD ? BF ? BITMASKING ?
!!!!! SOLVE THE SUBTASKS FIRST: IT'S TOTALLY OK NOT TO SOLVE THE PROBLEM ENTIRELY
Search for multiple approaches: select the seemingly optimal one
Remember that you're the king of CP
Change your approach
Imagine Corner cases before submitting
Don't spend too much time on the problem: move on !
*/
