The program must accept an integer N as the input. For each digit D in N, the program must print the digit D for X times where X is formed by removing the digit D from the integer N as the output.

**Note: **The value of N is always not a multiple of 100.

**Boundary Condition(s):**

101 <= N< 10^5

**Input Format:**

The first line contains N.

**Output Format**:

The lines contain the integers separated by a space.

**Example Input/Output 1:**

Input

121

Output

111111111111111111111

22222222222

111111111111

**Explanation**

For the first digit in 121, the value of X is 21. So 1 is printed for 21 times. For the second digit in 121, the value of X is 11. So 2 is printed for 11 times. For the third digit in 121, the value of X is 12. So 1 is printed for 12 times.

**Example Input/Output 2:**

Input:

210

Output:

2222222222

11111111111111111111

000000000000000000000

```
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n;
scanf("%d",&n);
int a[1000000];
int k=0;
//storing each element in a array
while(n>0)
{
a[k]=n%10;
n/=10;
k++;
}
int m=k-1;
k--;
int f;
while(m>=0)
{
f=0;
//calculating the number without its position
for (int i=k;i>=0; i--)
{
if(i!=m)
f=(f*10)+a[i];
}
//printing the digit
for (int i=0;i<f;i++)
printf("%d ",a[m]);
printf("\n");
m--;
}
}
```