Help solving Python problem

[WHelped]

I am not writing and spiking English good, do not be so angry, please
There is a list of int and natural numbers k,m. Print the min i that is true a[i] + a[i+1] + ... + a[i + k] == m. If there is not such i then print 0. Don't use nested loops and more than one list.
Input:
n, k and m (m <= 10000, 0 < k < n <= 30000), n - number of elements in list. At the next raw elements |i|<100
Output:
Answer

One of the tests:
Input:
4 1 22
9 13 10 -11
Output:
1
My code:

n, k, m = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
rez=0
for i in range(n):
	if sum(a[i:i+k+1])==m:
		rez=i+1
		break
print(rez if rez else 0)

Code fails on the last of test , not the test I have shown!!!
Maybe You know what is the problem. If "yes",please write me!

[nilamo]

Code fails on the last of test Cry Arrow , not the test I have shown!!!

Do you know what test it fails for?

[WHelped]

Sory, I don't know. Thank you for reply

[scidam]

Try this:

def find_i(a, m, k):
    n = len(a)
    for i in range(n - k):
        if sum(a[i:i + k + 1]) == m: 
            return i
    return 0

[WHelped]

Try this:

def find_i(a, m, k):
    n = len(a)
    for i in range(n - k):
        if sum(a[i:i + k + 1]) == m: 
            return i
    return 0

I tried, but the programm failed at 5 test. Looks strange, didn't it?

---

Yes, I found it!!! This is the corect code:

n, k, m = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
rez=0
for i in range(n-k):
    if sum(a[i:i+k+1])==m:
        rez+=i+1
        break
print(rez if rez else 0)