# Sum of Digits

### Problem Statement :

```You're given an integer N. Write a program to calculate the sum of all the digits of N.

Input

The first line contains an integer T, the total number of testcases. Then follow T lines, each line contains an integer N.

Output

For each test case, calculate the sum of digits of N, and display it in a new line.

Constraints
1 ≤ T ≤ 1000
1 ≤ N ≤ 1000000

Example

Input

3
12345
31203
2123

Output

15
9
8```

### Solution :

```                            ```Solution in C :

#include <stdio.h>

int main(void) {
int t;
scanf("%d",&t);
while (t--){
int n,m,sum=0;
scanf ("%d",&n);
while(n>0){

m=n%10;
n=n/10;

sum=sum+m;
}
printf("%d\n",sum);

}

return 0;
}```
```

```                        ```Solution in C++ :

#include <iostream>
using namespace std;

int main()
{
int t;
cin>>t;

while(t--)
{
int n;
cin>>n;

int sum=0;

while(n>0)
{
sum+=(n%10);
n/=10;
}

cout<<sum<<"\n";

}
return 0;
}```
```

```                        ```Solution in Java :

import java.util.*;
class Main{
public static void main(String[] args){
try
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for(int i=0;i<n;i++){
int a = sc.nextInt();
int sum = 0 ;
while(a>0){
int temp = a%10;
sum +=temp;
a/=10;
}
System.out.println(sum);
}
}
catch(Exception e)
{
return;
}
}
}```
```

```                        ```Solution in Python :

T=int(input())
for i in range(T):
num=int(input())
sum=0
for j in str(num):
sum+=int(j)
print(sum)```
```

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