##### Short Problem Definition:

Animesh has *N* empty candy jars, numbered from 1 to *N*, with infinite capacity. He performs *M* operations. Each operation is described by 3 integers *a, b* and *k*. Here, *a* and *b* are indices of the jars, and *k* is the number of candies to be added inside each jar whose index lies between_a_ and *b* (both inclusive). Can you tell the average number of candies after *M* operations?

##### Link

##### Complexity:

time complexity is `O(N)`

space complexity is `O(1)`

##### Execution:

Keep a sum variable. Compute the average at the end.

##### Solution:

```
#!/usr/bin/py
if __name__ == '__main__':
n,m = map(int, raw_input().split())
answer = 0
for _ in xrange(m):
a, b, k = map(int, raw_input().split())
answer += (abs(a-b)+1)*k
print answer//n
```

```
#include<iostream>
#include<vector>
#include<math.h>
#include<array>
using namespace std;
int main()
{
long n,m;
long answer=0,t;
cin>>n>>m;
t = m;
while(t--){
long a,b,k;
cin>>a>>b>>k;
answer = answer + (abs(a-b)+1)*k;
}
answer = floor(answer/n);
cout<<answer<<endl;
return 0;
}
```