Educational Codeforces Round 87 (Rated for Div. 2) B. Ternary String（贪心？蛮好一题）

You are given a string ss such that each its character is either 1, 2, or 3. You have to choose the shortest contiguous substring of ss such that it contains each of these three characters at least once.

A contiguous substring of string ss is a string that can be obtained from ss by removing some (possibly zero) characters from the beginning of ss and some (possibly zero) characters from the end of ss.

Input

The first line contains one integer tt (1≤t≤200001≤t≤20000) — the number of test cases.

Each test case consists of one line containing the string ss (1≤|s|≤2000001≤|s|≤200000). It is guaranteed that each character of ss is either 1, 2, or 3.

The sum of lengths of all strings in all test cases does not exceed 200000200000.

Output

For each test case, print one integer — the length of the shortest contiguous substring of ss containing all three types of characters at least once. If there is no such substring, print 00 instead.

Example

Input

Copy

7
123
12222133333332
112233
332211
12121212
333333
31121

Output

Copy

3
3
4
4
0
0
4
输出最短的同时含有1 2 3这三个数字的子串的长度。
遍历字符串，用三个变量pos[1] pos[2] pos[3]表示在当前位置之前最近一次出现1 2 3的位置（初始化为-1）。
每遍历到一个位置先更新对应的pos值，如果此时三个变量都不为-1的话更新答案即可。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        string s;
        cin>>s;
        int pos[4],i;//pos1代表离着当前位置最近的1的位置 其他同理 
        int ans=0x3f3f3f3f;
        pos[1]=pos[2]=pos[3]=-1;
        for(i=0;i<s.size();i++)
        {
            if(i==0)
            {
                pos[s[i]-'0']=0;
                continue;
            }
            int now=s[i]-'0';
            pos[now]=i;
            if(pos[1]!=-1&&pos[2]!=-1&&pos[3]!=-1)
            {
                if(now==1) ans=min(ans,i-min(pos[2],pos[3])+1);
                else if(now==2) ans=min(ans,i-min(pos[1],pos[3])+1);
                else if(now==3) ans=min(ans,i-min(pos[1],pos[2])+1);
            }
        }
        if(ans==0x3f3f3f3f)cout<<0<<endl;
        else cout<<ans<<endl;
    }
    return 0;
}