Problem:
Score: 200 points

Problem Statement

Today, the memorable AtCoder Beginner Contest 100 takes place. On this occasion, Takahashi would like to give an integer to Ringo.

As the name of the contest is AtCoder Beginner Contest 100, Ringo would be happy if he is given a positive integer that can be divided by 100 exactly D times.

Find the N-th smallest integer that would make Ringo happy.

Constraints

D is 0, 1 or 2.

N is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

D N

Output

Print the N-th smallest integer that can be divided by 100 exactly D times.

Sample Input 1

0 5

Sample Output 1

5

The integers that can be divided by 100 exactly 0 times (that is, not divisible by 100) are as follows: 1, 2, 3, 4, 5, 6, 7, ...

Thus, the 5-th smallest integer that would make Ringo happy is 5.

Sample Input 2

1 11

Sample Output 2

1100

The integers that can be divided by 100 exactly once are as follows: 100, 200, 300, 400, 500, 600, 700, 800, 900, 1 \ 000, 1 \ 100, ...

Thus, the integer we are seeking is 1 \ 100.

Sample Input 3

2 85

Sample Output 3

850000

The integers that can be divided by 100 exactly twice are as follows: 10 \ 000, 20 \ 000, 30 \ 000, ...

Thus, the integer we are seeking is 850 \ 000.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <cstdlib>
#include <iomanip>
#include <cmath>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdlib>
#include <limits>
#include <numeric>
#include <queue>
#include <set>
#include <bitset>
#include <map>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<char> vc;
typedef queue<int, int> qii;
typedef pair<int, int> pii;

#define PI 3.14159265359
#define rep(i,a,b) for(int i=a;i<b;i++)
#define repz(i,n) rep(i,0,n)

//int型を受け取ってその桁数を返す
int check_digits(int n) {
	int digits = 0;
	while (n != 0) {
		n /= 10;
		digits++;
	}
	return digits;
}


//最大公約数を返す関数
int gcd(int n, int m) {
	while (n != 0) {
		if (n < m) swap(n, m);
		n = n % m;
	}
	return m;
}


//2点間の距離を返す
float dis(int x1, int y1, int x2, int y2) {
	return sqrt(pow(abs(x1 - x2), 2) + pow(abs(y1 - y2), 2));
}


int main() {
	int d, n;
	cin >> d >> n;
	
	if (n == 100)cout << 101 * pow(100, d) << endl;
	else cout << n * pow(100, d) << endl;
	
	system("pause");
	return 0;
}