题干：

Vasya came up with a password to register for EatForces — a string ss. The password in EatForces should be a string, consisting of lowercase and uppercase Latin letters and digits.

But since EatForces takes care of the security of its users, user passwords must contain at least one digit, at least one uppercase Latin letter and at least one lowercase Latin letter. For example, the passwords "abaCABA12", "Z7q" and "3R24m" are valid, and the passwords "qwerty", "qwerty12345" and "Password" are not.

A substring of string ss is a string x=slsl+1…sl+len−1(1≤l≤|s|,0≤len≤|s|−l+1)x=slsl+1…sl+len−1(1≤l≤|s|,0≤len≤|s|−l+1). lenlen is the length of the substring. Note that the empty string is also considered a substring of ss, it has the length 00.

Vasya's password, however, may come too weak for the security settings of EatForces. He likes his password, so he wants to replace some its substring with another string of the same length in order to satisfy the above conditions. This operation should be performed exactly once, and the chosen string should have the minimal possible length.

Note that the length of ss should not change after the replacement of the substring, and the string itself should contain only lowercase and uppercase Latin letters and digits.

Input

The first line contains a single integer TT (1≤T≤1001≤T≤100) — the number of testcases.

Each of the next TT lines contains the initial password s (3≤|s|≤100)s (3≤|s|≤100), consisting of lowercase and uppercase Latin letters and digits.

Only T=1T=1 is allowed for hacks.

Output

For each testcase print a renewed password, which corresponds to given conditions.

The length of the replaced substring is calculated as following: write down all the changed positions. If there are none, then the length is 00. Otherwise the length is the difference between the first and the last changed position plus one. For example, the length of the changed substring between the passwords "abcdef" →→"a7cdEf" is 44, because the changed positions are 22 and 55, thus (5−2)+1=4(5−2)+1=4.

It is guaranteed that such a password always exists.

If there are several suitable passwords — output any of them.

Example

Input

2
abcDCE
htQw27

Output

abcD4E
htQw27

Note

In the first example Vasya's password lacks a digit, he replaces substring "C" with "4" and gets password "abcD4E". That means, he changed the substring of length 1.

In the second example Vasya's password is ok from the beginning, and nothing has to be changed. That is the same as replacing the empty substring with another empty substring (length 0).

解题报告：

 

错误代码1：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
string s;
int main() 
{int t;cin>>t;while(t--) {cin>>s;int low=-1,upp=-1,dig=-1;int l=0,u=0,d=0;for(int i = 0; i<s.length(); i++) {if(s[i] >= 'a' && s[i] <= 'z') low = i,l++;if(s[i] >= 'A' && s[i] <= 'Z') upp = i,u++;if(s[i] >= '1' && s[i] <= '9') dig = i,d++; }if(low == -1) {if(u>=2) s[upp] = 'a';else s[dig] = 'a';}if(upp == -1) {if(l>=2) s[low] = 'A';else s[dig] = 'A';}if(dig == -1) {if(l>=2) s[low] = '1';else s[upp] = '1';}cout << s << endl;}return 0;
}

1wa在了思路上（代码如上），这样写是不对的，因为有可能upp和dig都等于-1，所以都需要low进行更新，举个例子“aaa”，就会发现样例结果是错的。

2wa在了判断数字，，，应该是s[i]>='0' 为甚要写>=‘1’    怎么想的？？？

AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
char s[505];
int l[505],u[505],d[505];
int cnt1,cnt2,cnt3;
int main() 
{int t;cin>>t;getchar();while(t--) {gets(s);int len = strlen(s);int low=-1,upp=-1,dig=-1;cnt1=cnt2=cnt3=0;for(int i = 1; i<505; i++) l[i]=u[i]=d[i]=0;for(int i = 0; i<len; i++) {if(s[i] >= 'a' && s[i] <= 'z') low = i,l[++cnt1] = i;if(s[i] >= 'A' && s[i] <= 'Z') upp = i,u[++cnt2] = i;if(s[i] >= '0' && s[i] <= '9') dig = i,d[++cnt3] = i; }if(low == -1) {if(cnt2 >= 2) s[u[cnt2--]]='a';else s[d[cnt3--]] = 'a';}if(upp == -1) {if(cnt1>=2) s[l[cnt1--]] = 'A';else s[d[cnt3--]] = 'A';}if(dig == -1) {if(cnt1>=2) s[l[cnt1--]] = '1';else s[u[cnt2--]] = '1';}cout << s << endl;}return 0;
}

发现了别人的优秀代码：嗯还是有很多可以学习的。。要善于运用vector等容器啊

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using vpii = vector<pii>;
using vpll = vector<pll>;
template <typename T>
using ost = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define forn(i,n) for(int i=0; i<int(n); ++i)
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define ff first
#define ss secondint main() {ios::sync_with_stdio(false);int T;cin >> T;while (T--) {string s;cin >> s;bool lo=0, hi=0, dig=0;vi x;int n=s.size();for (int i=0; i<n; i++) {if (isdigit(s[i])) {if (dig) x.pb(i);dig=1;}else if (islower(s[i])) {if (lo) x.pb(i);lo=1;}else {if (hi) x.pb(i);hi=1;}}if (!dig) {s[x.back()]='1';x.pop_back();}if (!lo) {s[x.back()]='a';x.pop_back();}if (!hi) {s[x.back()]='A';x.pop_back();}cout << s << '\n';}
}