Problem:
Score : 400 points

Problem Statement

You are given integers N and M.

Consider a sequence a of length N consisting of positive integers such that a_1 + a_2 + ... + a_N = M. Find the maximum possible value of the greatest common divisor of a_1, a_2, ..., a_N.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

N \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the maximum possible value of the greatest common divisor of a sequence a_1, a_2, ..., a_N that satisfies the condition.

Sample Input 1

3 14

Sample Output 1

2

Consider the sequence (a_1, a_2, a_3) = (2, 4, 8). Their greatest common divisor is 2, and this is the maximum value.

Sample Input 2

10 123

Sample Output 2

3

Sample Input 3

100000 1000000000

Sample Output 3

10000

Solution in C (Clang 3.8.0) (judge verdict: Accepted)

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <inttypes.h>
#include <ctype.h>
#include <stdint.h>
#include <string.h>
#include <wchar.h>

#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define ABS(a) ((a) < 0 ? -(a) : (a))
#define ABSS(a, b) ((a) > (b) ? (a) - (b) : (b) - (a))

int compare_f(const void *a, const void *b) {
	return *((size_t *)a) < *((size_t *)b) ? -1 : 1;
}

static size_t comb(const size_t n, const size_t r) {
	size_t result = 1;

	for (size_t i = 0; i < r; i++) {
		result *= n - i;
		result /= i + 1;
	}

	return result;
}

int main(void) {
	size_t N, M;

	scanf("%zu %zu", &N, &M);

	size_t commonDivisor = M / N;

	while (1 <= commonDivisor) {
		if (M%commonDivisor == 0) {
			break;
		}

		commonDivisor--;
	}

	printf("%zu\n", commonDivisor);

	return 0;
}