【Codeforces Round 339 (Div 2)B】【水题】Gena's Code 若干10数+1特殊数的乘积

B. Gena's Code

time limit per test

0.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It's the year 4527 and the tanks game that we all know and love still exists. There also exists Great Gena's code, written in 2016. The problem this code solves is: given the number of tanks that go into the battle from each country, find their product. If it is turns to be too large, then the servers might have not enough time to assign tanks into teams and the whole game will collapse!

There are exactly n distinct countries in the world and the i-th country added a_i tanks to the game. As the developers of the game are perfectionists, the number of tanks from each country is beautiful. A beautiful number, according to the developers, is such number that its decimal representation consists only of digits '1' and '0', moreover it contains at most one digit '1'. However, due to complaints from players, some number of tanks of one country was removed from the game, hence the number of tanks of this country may not remain beautiful.

Your task is to write the program that solves exactly the same problem in order to verify Gena's code correctness. Just in case.

Input

The first line of the input contains the number of countries n (1 ≤ n ≤ 100 000). The second line contains n non-negative integers a_iwithout leading zeroes — the number of tanks of the i-th country.

It is guaranteed that the second line contains at least n - 1 beautiful numbers and the total length of all these number's representations doesn't exceed 100 000.

Output

Print a single number without leading zeroes — the product of the number of tanks presented by each country.

Examples

input

35 10 1

output

50

input

41 1 10 11

output

110

input

50 3 1 100 1

output

0

Note

In sample 1 numbers 10 and 1 are beautiful, number 5 is not not.

In sample 2 number 11 is not beautiful (contains two '1's), all others are beautiful.

In sample 3 number 3 is not beautiful, all others are beautiful.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }const int N = 1e5+10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;int n;char s[N], t[N];int main(){while (~scanf("%d", &n)){t[0] = '1';bool ZERO = 0;int zeronum = 0;for (int i = 0; i < n; ++i){scanf("%s", s);if (s[0] == '0')ZERO = 1;int one = 0;int other = 0;int zero = 0;for (int i = 0; s[i]; ++i){if (s[i] == '1')++one;else if (s[i] == '0')++zero;else ++other;}if (other || one > 1)MC(t, s);else zeronum += zero;}if (ZERO)puts("0");else{printf("%s", t);for (int i = 1; i <= zeronum; ++i)printf("0");puts("");}}return 0;}/*【题意】我们定义"one-number"为只有0或1构成的数，且数中最多只有1个1。给你n（1e5）个数，这些数只有一个数不是one-number，这些数的位数之和可达1e5，问你这些数的乘积是多少。【类型】水题【分析】我们只要判下乘积是否为0.否则把非one-number移到首位，剩余的0放到尾端就好了。*/