Everyone knows that agents in Valorant decide, who will play as attackers, and who will play as defenders. To do that Raze and Breach decided to play t matches of a digit game...

In each of t matches of the digit game, a positive integer is generated. It consists of n digits. The digits of this integer are numerated from 1 to n from the highest-order digit to the lowest-order digit. After this integer is announced, the match starts.

Agents play in turns. Raze starts. In one turn an agent can choose any unmarked digit and mark it. Raze can choose digits on odd positions, but can not choose digits on even positions. Breach can choose digits on even positions, but can not choose digits on odd positions. The match ends, when there is only one unmarked digit left. If the single last digit is odd, then Raze wins, else Breach wins.

It can be proved, that before the end of the match (for every initial integer with n digits) each agent has an ability to make a turn, i.e. there is at least one unmarked digit, that stands on a position of required parity.

For each of t matches find out, which agent wins, if both of them want to win and play optimally.

Input

First line of input contains an integer t (1 ≤ t ≤ 100) — the number of matches.

The first line of each match description contains an integer n (1 ≤ n ≤ 10^3) — the number of digits of the generated number.

The second line of each match description contains an n-digit positive integer without leading zeros.

Output

For each match print 1, if Raze wins, and 2, if Breach wins.

Example

Input


4
1
2
1
3
3
102
4
2069


Output


2
1
1
2

Note

In the first match no one can make a turn, the only digit left is 2, it's even, so Breach wins.

In the second match the only digit left is 3, it's odd, so Raze wins.

In the third match Raze can mark the last digit, after that Breach can only mark 0. 1 will be the last digit left, it's odd, so Raze wins.

In the fourth match no matter how Raze plays, Breach can mark 9, and in the end there will be digit 0. It's even, so Breach wins.
"""
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"""


import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush, nsmallest
from math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
from decimal import Decimal
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(var, end="\n"): sys.stdout.write(str(var)+end)
def outa(*var, end="\n"): sys.stdout.write(' '.join(map(str, var)) + end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


for _ in range(int(data())):
    n = int(data())
    s = list(map(int, data()))
    odd = 2
    if n & 1:
        for i in range(0, n, 2):
            if s[i] & 1:
                odd = 1
                break
        out(odd)
        continue
    even = 1
    for i in range(1, n, 2):
        if not(s[i] & 1):
            even = 2
            break
    out(even)
