D. Print a 1337-string...(字符串构造)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You are given an integer n.

You have to find a sequence s consisting of digits {1,3,7} such that it has exactly n subsequences equal to 1337.

For example, sequence 337133377 has 6 subsequences equal to 1337:

1. 1. 
      
2. 3371_33_3_77_ (you can remove the second and fifth characters);
3. 3371_3_33_77_ (you can remove the third and fifth characters);
4. 3371_3_3_377_ (you can remove the fourth and fifth characters);
5. 3371_33_3_7_7 (you can remove the second and sixth characters);
6. 3371_3_33_7_7 (you can remove the third and sixth characters);
7. 3371_3_3_37_7 (you can remove the fourth and sixth characters).
   1. 
      

Note that the length of the sequence s must not exceed 105.

You have to answer t independent queries.

Input

The first line contains one integer t (1≤t≤10) — the number of queries.

Next t lines contains a description of queries: the i-th line contains one integer ni (1≤ni≤109).

Output

For the i-th query print one string si (1≤|si|≤105) consisting of digits {1,3,7}. String si must have exactly ni subsequences 1337. If there are multiple such strings, print any of them.

Example

input

Copy

2
6
1

output

Copy

113337
1337

因为n的范围小于1e9，题目要求字符串最长为1e5，那么不能只加1：11111337这样构造，会超长度。

此时可以发现添加3提供的答案是m*(m-1)/2，1e9，最多1e5个3就可以构造出来了。

于是暴力找出最小的3的个数，可此时m*（m-1）不一定等于n，那么就将m--，然后剩下的用一插在后面俩个三的前面单独提供答案就可以了，形式：133333333...111111....337。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    int _;
    cin>>_;
    while(_--)
    {
        int n;cin>>n;
        ll m;
        string ans="1";
        for(m=0;;m++)
        {
            if(m*(m-1)/2>n) break;
        }
        --m;
        //printf("m:%lld\n",m);
        ll n1=n-m*(m-1)/2;
        ll tmp=m;
        while(tmp--) ans+='3';
        if(n1)
        {
            ans.pop_back();
            ans.pop_back();
            while(n1--) ans+='1';
            ans+='3';
            ans+='3';
        }
        ans+='7';
        cout<<ans<<endl;
    }
}