Problem:
Score : 300 points

Problem Statement

We have N clocks. The hand of the i-th clock (1≤i≤N) rotates through 360° in exactly T_i seconds.

Initially, the hand of every clock stands still, pointing directly upward.

Now, Dolphin starts all the clocks simultaneously.

In how many seconds will the hand of every clock point directly upward again?

Constraints

1≤N≤100

1≤T_i≤10^{18}

All input values are integers.

The correct answer is at most 10^{18} seconds.

Input

Input is given from Standard Input in the following format:

N
T_1
:
T_N

Output

Print the number of seconds after which the hand of every clock point directly upward again.

Sample Input 1

2
2
3

Sample Output 1

6

We have two clocks. The time when the hand of each clock points upward is as follows:

Clock 1: 2, 4, 6, ... seconds after the beginning

Clock 2: 3, 6, 9, ... seconds after the beginning

Therefore, it takes 6 seconds until the hands of both clocks point directly upward.

Sample Input 2

5
2
5
10
1000000000000000000
1000000000000000000

Sample Output 2

1000000000000000000

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

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

// 内部変数 - テスト用
#ifdef D_TEST
	static int siRes;
	static FILE *szpFpT, *szpFpA;
#endif

// 最大公約数取得
long long
fGetGcd(
	long long llVal1			// <I> 値１
	, long long llVal2			// <I> 値２
)
{
	long long llWork;

	// ユークリッドの互除法
	if (llVal1 < llVal2) {
		llWork = llVal1;
		llVal1 = llVal2;
		llVal2 = llWork;
	}
	while (llVal2 > 0) {
		llWork = llVal1 % llVal2;
		llVal1 = llVal2;
		llVal2 = llWork;
	}

	return llVal1;
}

// 最小公倍数取得
long long
fGetLcm(
	long long llVal1			// <I> 値１
	, long long llVal2			// <I> 値２
)
{
	// 最大公約数取得
	long long llGcd = fGetGcd(llVal1, llVal2);

	// 最小公倍数取得
	return llVal1 / llGcd * llVal2;
}

// 実行メイン
int
fMain(
	int piTNo					// <I> テスト番号 1～
)
{
	int i;
	char lc1Buf[1024], lc1Out[1024];

	// テストファイルオープン
#ifdef D_TEST
	sprintf(lc1Buf, ".\\Test\\T%d.txt", piTNo);
	szpFpT = fopen(lc1Buf, "r");
	sprintf(lc1Buf, ".\\Test\\A%d.txt", piTNo);
	szpFpA = fopen(lc1Buf, "r");
	siRes = 0;
#endif

	// 時計数取得
	int liCCnt;
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif
	sscanf(lc1Buf, "%d", &liCCnt);

	// １台目取得
	long long llSec;
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
	fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif
	sscanf(lc1Buf, "%lld", &llSec);

	// ２台目以降取得
	for (i = 1; i < liCCnt; i++) {
		long long llSec2;
#ifdef D_TEST
		fgets(lc1Buf, sizeof(lc1Buf), szpFpT);
#else
		fgets(lc1Buf, sizeof(lc1Buf), stdin);
#endif
		sscanf(lc1Buf, "%lld", &llSec2);

		// 最小公倍数取得
		llSec = fGetLcm(llSec, llSec2);
	}

	// 結果セット
	sprintf(lc1Out, "%lld\n", llSec);

	// 結果表示
#ifdef D_TEST
	fgets(lc1Buf, sizeof(lc1Buf), szpFpA);
	if (strcmp(lc1Buf, lc1Out)) {
		siRes = -1;
	}
#else
	printf("%s", lc1Out);
#endif

	// テストファイルクローズ
#ifdef D_TEST
	fclose(szpFpT);
	fclose(szpFpA);
#endif

	// テスト結果
#ifdef D_TEST
	if (siRes == 0) {
		printf("OK %d\n", piTNo);
	}
	else {
		printf("NG %d\n", piTNo);
	}
#endif

	return 0;
}

int
main()
{

#ifdef D_TEST
	int i;
	for (i = D_TEST_SNO; i <= D_TEST_ENO; i++) {
		fMain(i);
	}
#else
	fMain(0);
#endif

	return 0;
}