求每一种字符出现次数的前缀和，然后二分答案的长度就可
#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()
const LL N=2e5+10;
string s;
int cnt[N][3]; //求前缀和，然后二分
int binarySearch(){
int left=3;
int right=s.length();
int mid=0;
int ans=0;
while(left<=right){
mid=(left+right)>>1;
bool flag=false;
rep(i,0,s.length()-mid+1){
if(cnt[i+mid][0]-cnt[i][0]>0&&cnt[i+mid][1]-cnt[i][1]>0
&&cnt[i+mid][2]-cnt[i][2]>0){
flag = true;
break;
}
}
if(flag){
ans=mid;
right=mid-1;
}else{
left=mid+1;
}
}
return ans;
}
int main(int argc, char const *argv[])
{
// #define DEBUG
#ifdef DEBUG
freopen("1.dat","r",stdin);
freopen("ans.dat","w",stdout);
#endif
LL _;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>_;
while(_--){
s="";
cin>>s;
if(s.length()<3){
cout<<0<<endl;
continue;
}
cnt[0][0]=cnt[0][1]=cnt[0][2]=0;
rep(i,0,s.length()){
rep(j,0,3)
cnt[i+1][j]=cnt[i][j];
cnt[i+1][s[i]-'1']++;
}
cout<<binarySearch()<<endl;
}
return 0;
}