This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Input

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

Output

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

Sample Input

21
85
789
0

Sample Output

21 0
85 5
789 62 思路:打表求出H-prime,再两两相乘,用树状数组优化求和问题即可
typedef long long LL;
typedef pair<LL, LL> PLL; const int maxm = 1e6+; bool prime[maxm];
int vis[maxm];
int jud[maxm], siz = , C[maxm]; void add(int x, int val) {
for(; x < maxm; x += lowbit(x))
C[x] += val;
} LL getsum(int x) {
LL ret = ;
for(; x; x -= lowbit(x))
ret += C[x];
return ret;
} void getHprime() {
for(int i = ; i < maxm; i += ) {
if(!prime[i]) {
for(int j = *i; j < maxm; j += i)
prime[j] = true;
jud[siz++] = i;
for(int k = ; k < siz; ++k) {
if(maxm / i >= jud[k]) {
if(!vis[jud[k] * i]++)
add(i*jud[k], );
} else
break;
} }
} } int main() {
getHprime();
int n;
while(scanf("%d", &n) && n) {
printf("%d %lld\n", n, getsum(n));
}
return ;
}
												

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