#include<bits/stdc++.h>
typedef long long ll;
#define mod 1000000007
#define all(x) x.begin(),x.end()
ll fpow(ll n, ll k, ll p = mod) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
ll inv(int a, ll p = mod) {return fpow(a, p - 2, p);}
using namespace std;
ll chk[26];	//hash array
ll dp1[26][100003],dp2[26][100003];
ll n;
string s;
void init1()
{
	memset(chk,0,sizeof(chk));
	for(ll i=0;s[i]!='\0';i++)
		chk[s[i]-'a']++;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	ll t;cin>>t;
	while(t--)
	{
		cin>>s;
		n=s.length();
		init1();		//hashes the characters of string s in a frequency array
		ll ini_cnt=0;	//ini_cnt stores the number of pairs (i,j) such that i<j and s[i]<s[j] in the initial string
		//this is calculated using hashing the characters in an array of sie 26.
		//we iterate from the last characters, add to ini_cnt all hash values of characters less
		//than the current character
		for(ll i=s.length()-1;i>=0;i--)
		{
			ll temp=s[i]-'a';
			for(ll j=0;j<temp;j++)
			{
				ini_cnt+=chk[j];
			}
			chk[temp]--;
		}

		//cout<<"ini ans is "<<ini_cnt<<endl;
		//2 dp matrices
		//dp1[i][j] stores how many characters from the start uptil position j are less than i
		//dp2[i][j] stores how many characters from the end uptil position j are greater than i 
		
		memset(dp1,0,sizeof(dp1));
		memset(dp2,0,sizeof(dp2));

		//calculating dp1
		for(ll i=0;i<26;i++)
		{
			if(s[0]-'a'<i)
				dp1[i][0]=1;

			for(ll j=1;j<n;j++)
			{
				ll temp=s[j]-'a';
				if(temp<i)
					dp1[i][j]=dp1[i][j-1]+1;
				else
					dp1[i][j]=dp1[i][j-1];
			}
		}

		//calculating dp2
		for(ll i=0;i<26;i++)
		{
			if(s[s.length()-1]-'a'>i)
				dp2[i][n-1]=1;

			for(ll j=n-2;j>=0;j--)
			{
				ll temp=s[j]-'a';
				if(temp>i)
					dp2[i][j]=dp2[i][j+1]+1;
				else
					dp2[i][j]=dp2[i][j+1];
			}
		}

		//ans stores our final answer
		ll ans=LLONG_MAX;
		for(ll i=0;i<n;i++)
		{
			ll tcnt=ini_cnt;
			if(i==0)
			{
				tcnt-=dp2[s[i]-'a'][i+1];
				for(ll j=0;j<26;j++)
				{
					ll x=abs(j-(s[i]-'a'));
					ll temp=dp2[j][i+1];
					ans=min(ans,x+temp+tcnt);
				}
			}
			else if(i==n-1)
			{
				tcnt-=dp1[s[i]-'a'][i-1];
				for(ll j=0;j<26;j++)
				{
					ll x=abs(j-(s[i]-'a'));
					ll temp=dp1[j][i-1];
					ans=min(ans,x+temp+tcnt);
				}
			}
			else
			{
				//we replace the i'th character by all 26 characters and calculate X and Y for each
				//X will be simply absolute value of the difference of their ascii values
				//Y will be calculated as follows:
				//tcnt stores the total count of pairs, when we replace the i'th character
				//we must subtract from tcnt the number of pairs that it contributed to
				//these are calculated as sum of :
				//number of alphabets smaller than i'th character till index i-1
				//number of alphabets greater than i'th character till index i+1
				//and are replace by respective number for the replaced character
				//finally we take the minimum of all possible configurations
				tcnt-=dp1[s[i]-'a'][i-1];
				tcnt-=dp2[s[i]-'a'][i+1];
				for(ll j=0;j<26;j++)
				{
					ll x=abs(j-(s[i]-'a'));
					ll temp=dp1[j][i-1]+dp2[j][i+1];
					ans=min(ans,x+temp+tcnt);
				}
			}
		}

		cout<<ans<<endl;
	}	
}