**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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