I - Por Costel and the Pairs Gym - 100923I _思维啊——可惜我现在还没

We don’t know how Por Costel the pig arrived at FMI’s dance party. All we know is that he did.

The dance floor is hot because Por Costel broke the air conditioner. On the dance floor, the boys and girls are clumsy. There are boys at the party. The -th one has clumsiness level . There ale also girls at the party. The -th girl has clumsiness level as well. Por Costel finds that a pair of a boy and a girl can dance only if the product of their clumsiness levels is at most , i.e. girl clumsiness level * boy clumsiness level . Por Costel thinks that a sack of oats with a slice of leek could make a better dancing pair than the people at this party. Nonetheless, he would like to find out how many pairs (boy, girl) can dance. The input will contain the number representing the number of tests. The next lines will contain the number .

Input

The input file perechi3.in will contain the number representing the number of tests. The next lines will each contain a single integer .

Output

The output file perechi3.out should have lines. Line should hold the answer for the -th test.

Example

input

11

1

4

5

10

100

99

16

64

49

50

48

output

1

8

10

27

482

473

50

280

201

207

198

题意： 有n个男的，n个女的，第i个人都有为当前一个大小为i的懒惰值，当一男一女懒惰值的乘积<=n他们就就可以一起跳舞，请问有多少种组合可能；

解：

第i个男人可以和前n/i个女人跳舞，女的也是这样，会有i个重复，但会多减一个坐标相同的加上就好

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
freopen("perechi3.in","r",stdin);
freopen("perechi3.out","w",stdout);
ll t,n,i,j;
cin>>t;
{
while(t--)
{
scanf("%lld",&n);
ll ans=0,pos=0;
for(i=1;i*i<=n;i++)
{
ans+=2*(n/i-i)+1;
}
printf("%lld\n", ans);
}
}
return 0;
}