B. Mislove Has Lost an Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mislove had an array a1a1, a2a2, ⋯⋯, anan of nn positive integers, but he has lost it. He only remembers the following facts about it:

• The number of different numbers in the array is not less than ll and is not greater than rr;
• For each array's element aiai either ai=1ai=1 or aiai is even and there is a number ai2ai2 in the array.

For example, if n=5n=5, l=2l=2, r=3r=3 then an array could be [1,2,2,4,4][1,2,2,4,4] or [1,1,1,1,2][1,1,1,1,2]; but it couldn't be [1,2,2,4,8][1,2,2,4,8] because this array contains 44 different numbers; it couldn't be [1,2,2,3,3][1,2,2,3,3] because 33 is odd and isn't equal to 11; and it couldn't be [1,1,2,2,16][1,1,2,2,16]because there is a number 1616 in the array but there isn't a number 162=8162=8.

According to these facts, he is asking you to count the minimal and the maximal possible sums of all elements in an array.

Input

The only input line contains three integers nn, ll and rr (1≤n≤10001≤n≤1000, 1≤l≤r≤min(n,20)1≤l≤r≤min(n,20)) — an array's size, the minimal number and the maximal number of distinct elements in an array.

Output

Output two numbers — the minimal and the maximal possible sums of all elements in an array.

Examples

input

Copy

4 2 2

output

Copy

5 7

input

Copy

5 1 5

output

Copy

5 31

Note

In the first example, an array could be the one of the following: [1,1,1,2][1,1,1,2], [1,1,2,2][1,1,2,2] or [1,2,2,2][1,2,2,2]. In the first case the minimal sum is reached and in the last case the maximal sum is reached.

In the second example, the minimal sum is reached at the array [1,1,1,1,1][1,1,1,1,1], and the maximal one is reached at the array [1,2,4,8,16][1,2,4,8,16].

解题说明：此题是一道数学题，最小的sum只需要完成元素种类个数是l，其他的用1去填。最大的sum，先完成元素种类个数是r，其他的用(2^(r-1)) 填。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

int main()
{
	int n, l, r, s_, s1_;
	scanf("%d %d %d", &n, &l, &r);
	if (l > 20)
	{
		l = 20;
	}
	if (r > 20)
	{
		r = 20;
	}
	s_ = (pow(2, l) - 1) + (n - l);
	s1_ = (pow(2, r) - 1) + (pow(2, r - 1) * (n - r));
	printf("%d %d\n", s_, s1_);
	return 0;
}