Chef published a blog post, and is now receiving many queries about it. On day $i$, he receives $Q_i$ queries. But Chef can answer at most $k$ queries in a single day. 
Chef always answers the maximum number of questions that he can on any given day (note however that this cannot be more than $k$). The remaining questions (if any) will be carried over to the next day.
Fortunately, after $n$ days, the queries have stopped. Chef would like to know the first day during which he has some free time, i.e. the first day when he answered less than $k$ questions. 

-----Input:-----
- First line will contain $T$, the number of testcases. Then the testcases follow.
- The first line of each testcase contains two space separated integers $n$ and $k$.
- The second line of each testcase contains $n$ space separated integers, namely $Q_1, Q_2, ... Q_n$.

-----Output:-----
For each testcase, output in a single line the first day during which chef answers less than $k$ questions. 

-----Constraints-----
- $1 \leq T \leq 10^5$
- $1 \leq $ sum of $n$ over all testcases $ \leq 10^5$
- $1 \leq k \leq 10^8$
- $0 \leq Q_i \leq 10^8$

-----Subtasks-----
- Subtask 1 - 20% points - Sum of $Q_i$ over all testcases and days $\leq 3 . 10^6$
- Subtask 2 - 80% points - Original constraints

-----Sample Input:-----
2 
6 5 
10 5 5 3 2 1 
1 1
100

-----Sample Output:-----
6
101

-----Explanation:-----
Test Case 1
On the first day, chef answers 5 questions and leaves the remaining 5 (out of the 10) for the future days.
On the second day, chef has 10 questions waiting to be answered (5 received on the second day and 5 unanswered questions from day 1). Chef answers 5 of these questions and leaves the remaining 5 for the future.
On the third day, chef has 10 questions waiting to be answered (5 received on the third day and 5 unanswered questions from earlier). Chef answers 5 of these questions and leaves the remaining 5 for later.
On the fourth day, chef has 8 questions waiting to be answered (3 received on the fourth day and 5 unanswered questions from earlier). Chef answers 5 of these questions and leaves the remaining 3 for later.
On the fifth day, chef has 5 questions waiting to be answered (2 received on the fifth day and 3 unanswered questions from earlier). Chef answers all 5 of these questions.
On the sixth day, chef has 1 question, which he answers. This is the first day he answers less than 5 questions, and so the answer is 6.
Test Case 2
Chef answers 1 question a day for the first 100 days. On day 101, he is free.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = '# cook your dish here\nt = int(input())\nwhile(t > 0):\n n,k = map(int,input().split())\n q = [int(x) for x in input().split()]\n carry = 0\n flag = 0\n for i in range(n):\n  if(q[i]+carry>=k):\n   carry = q[i] + carry - k\n  else:\n   flag = 1\n   print(i+1)\n   break\n if(carry > 0 and flag != 1):\n  rem = carry//k + 1 + n\n  print(rem)\n elif(carry == 0 and flag != 1):\n  print(n+1)\n t -= 1'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13199