A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1\cdot 6=6 \cdot 1=2\cdot 3=3\cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=ab and n=ban=ba are considered different if a \not = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating \sum_{i = 1}^nf(i)∑i=1n​f(i).

Input

The first line contains an integer T(T\le 20)T(T≤20), denoting the number of test cases.

For each test case, there first line has a integer n(n \le 2\cdot 10^7)n(n≤2⋅107).

Output

For each test case, print the answer \sum_{i = 1}^n f(i)∑i=1n​f(i).

Hint

\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18​f(i)=f(1)+⋯+f(8) =1+2+2+1+2+4+2+0=14=1+2+2+1+2+4+2+0=14.

樣例輸入複製

2
5
8

樣例輸出複製

8
14

題目來源

思路: 相同因子個數不超過2的數稱為squarefree（SF），F【i】= A*B的種數，AB都是SF，AB不相等的時候AB 和BA算兩種方案,求1~n內F【i】字首和。

魔改線性篩，太6了，Orz..................

程式碼：

#include<bits/stdc++.h>
using namespace std;
const int  maxn=2e7+5;
long long ans[maxn];
int tot,prime[maxn],vis[maxn],f[maxn];///prime儲存遍歷到的素數

void doit(){
    memset(vis,0,sizeof(vis));
    memset(prime,0,sizeof(prime));
    memset(f,0,sizeof(f));

    prime[1]=1;
    ans[1]=1;
    tot=0;

    for(int i=2;i<maxn;i++){
        if(!vis[i]){
            prime[tot++]=i;
            f[i]=2;
        }

        for(int j=0;j<tot&&i*prime[j]<maxn;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                if(i%(prime[j]*prime[j])==0) f[i*prime[j]]=0;///大於等於三次方GG
                else f[i*prime[j]]=f[i]/2;///等於二次方得變一半，相當在原基礎上少了另一個素數
                break;
            }
            else f[prime[j]*i]=f[i]*2;
        }
        ans[i]=ans[i-1]+f[i];
    }
}



int main(){
    doit();

    int t;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        printf("%lld\n",ans[n]);
    }
}