In the Republic of Remoteland, the people celebrate their independence day every year. However, as it was a long long time ago, nobody can remember when it was exactly. The only thing people can remember is that today, the number of days elapsed since their independence (D) is a perfect square, and moreover it is the largest possible such number one can form as a product of distinct numbers less than or equal to n.
As the years in Remoteland have 1,000,000,007 days, their citizens just need D modulo 1,000,000,007. Note that they are interested in the largest D, not in the largest D modulo 1,000,000,007.
Input
Every test case is described by a single line with an integer n, (1<=n<=10,000, 000). The input ends with a line containing 0.
Output
For each test case, output the number of days ago the Republic became independent, modulo 1,000,000,007, one per line.
Sample Input
4
9348095
6297540
0
Sample Output
4
177582252
644064736
題意：從[1,n]之間選擇若干個數字，使其為平方數，現在求最大的這個平方數。

分析： 從算數基本定理來考慮， 如果一個數字是平方數那麼其所有質因子的指數都是偶數，從這一點來考慮，[1,n]之間最大的我們首選n! ， 然後我們只需要列舉n!的所有的質因子，看其質因子的指數，如果是偶數，我們不用管，但是如果是奇數，我們就要除去一個當前質因子(因為如果有當前的這個質因子的話，那麼[1,n]之間一定有一個數字正好等於當前的這個質因子，所以我們除掉一個，就相當於將這個單獨的數去掉，不會影響什麼)。

程式碼

 #include<algorithm>
 #include<stdio.h>
 #define LL long long 
 using namespace std;

const
 
 int MAXN = 1e7+11;
const int MAXM = 1e7+11;
const int mod =  1e9+7;
const int inf = 0x3f3f3f3f;


int prm[MAXN+2],sz; bool su[MAXN+2]; int fac[MAXM];  
void init(){
    su[0]=su[1]=true;
    for(int i=2;i<=MAXN;i++){
        if(!su[i])  prm[++sz]=i;
        for(int j=1;j<=sz;j++){
            int
 
 t=i*prm[j];
            if(t>MAXN) break;
            su[t]=true;
            if(i%prm[j]==0) break;
        }
    }
    fac[0]=fac[1]=1; 
    for(int i=2;i<=MAXM;i++){
        fac[i]=(LL)fac[i-1]*i%mod;
    }
}
LL power(LL a,LL b,LL c){
    LL s=1,base=a%c;
    while(b){
        if(b&1) s=s*base%c;
        base=base*base%c;
        b>>=1;
    }
    return s;
}
LL inv(LL a){  // 費馬小定理求逆元
    return power(a,mod-2,mod);
}
void solve(LL n){
    LL ans=fac[n]; LL temp=1;
    for(int i=1;i<=sz;i++){
        LL cnt=0; LL t=n;
        if(n/prm[i]){
            while(t){
                cnt+=t/prm[i];
                t/=prm[i];
            } 
            if(cnt&1) temp=temp*prm[i]%mod;  // 將所有要除的先都存起來，最後求一下逆元， 一開始這裡直接就求逆元了，無限TLE。
        }else break; 
    }
    ans=ans*inv(temp)%mod; // 只要這裡求一次逆元就行了 
    printf("%lld\n",ans);
}
int main(){

     LL n; init();
     //printf("%d\n",sz);
     while(scanf("%lld",&n)&&n){
          solve(n);
    }
    return 0;
}