Problem:
Equal Range

For a given sequence $A = \{a_0, a_1, ..., a_{n-1}\}$ which is sorted by ascending order, print a pair of the lower bound and the upper bound for a specific value $k$ given as a query.

lower bound: the place pointing to the first element greater than or equal to a specific value, or $n$ if there is no such element.

upper bound: the place pointing to the first element greater than a specific value, or $n$ if there is no such element.

Input

The input is given in the following format.

$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$

The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.

Output

For each query, print the pari of the lower bound $l$ ($l = 0, 1, ..., n$) and upper bound $r$ ($r = 0, 1, ..., n$) separated by a space character in a line.

Constraints

$1 \leq n \leq 100,000$

$1 \leq q \leq 200,000$

$0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$

$0 \leq k_i \leq 1,000,000,000$

Sample Input 1

4
1 2 2 4
3
2
3
5

Sample Output 1

1 3
3 3
4 4

Solution in C (judge verdict: Accepted)

#include <stdio.h>

int a[100000];

int lower_bound(int lb, int ub, int k) {
	int m;
	while (ub - lb > 1) {
		m = (lb + ub) >> 1;
		if (a[m] < k) {
			lb = m;
		}
		else {
			ub = m;
		}
	}
	return ub;
}

int upper_bound(int lb, int ub, int k) {
	int m;
	while (ub - lb > 1) {
		m = (lb + ub) >> 1;
		if (a[m] <= k) {
			lb = m;
		}
		else {
			ub = m;
		}
	}
	return ub;
}

int main(void)
{
	int n, q, k, i;
	scanf("%d", &n);
	for (i = 0; i < n; i++) {
		scanf("%d", a + i);
	}
	scanf("%d", &q);
	while (q--) {
		scanf("%d", &k);
		printf("%d %d\n", lower_bound(-1, n, k), upper_bound(-1, n, k));
	}
	return 0;
}