D. Ticket Game

链接：https://codeforces.com/contest/1215/problem/D

Monocarp and Bicarp live in Berland, where every bus ticket consists of nn digits (nn is an even number). During the evening walk Monocarp and Bicarp found a ticket where some of the digits have been erased. The number of digits that have been erased is even.

Monocarp and Bicarp have decided to play a game with this ticket. Monocarp hates happy tickets, while Bicarp collects them. A ticket is considered happy if the sum of the first n2n2 digits of this ticket is equal to the sum of the last n2n2 digits.

Monocarp and Bicarp take turns (and Monocarp performs the first of them). During each turn, the current player must replace any erased digit with any digit from 00 to 99. The game ends when there are no erased digits in the ticket.

If the ticket is happy after all erased digits are replaced with decimal digits, then Bicarp wins. Otherwise, Monocarp wins. You have to determine who will win if both players play optimally.

Input

The first line contains one even integer nn (2≤n≤2⋅105)(2≤n≤2⋅105) — the number of digits in the ticket.

The second line contains a string of nn digits and "?" characters — the ticket which Monocarp and Bicarp have found. If the ii-th character is "?", then the ii-th digit is erased. Note that there may be leading zeroes. The number of "?" characters is even.

Output

If Monocarp wins, print "Monocarp" (without quotes). Otherwise print "Bicarp" (without quotes).

Examples

input

Copy

4
0523

output

Copy

Bicarp

input

Copy

2
??

output

Copy

Bicarp

input

Copy

8
?054??0?

output

Copy

Bicarp

input

Copy

6
???00?

output

Copy

Monocarp

Note

Since there is no question mark in the ticket in the first example, the winner is determined before the game even starts, and it is Bicarp.

In the second example, Bicarp also wins. After Monocarp chooses an erased digit and replaces it with a new one, Bicap can choose another position with an erased digit and replace it with the same digit, so the ticket is happy.

思路：

博弈题，可以简化成一个博弈模型

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
const int maxn=1e6+5;
long long n,s1=0,s2=0,b1=0,b2=0;
char a[200001];
int main()
{
	cin>>n;
	cin>>a;
	for(int i=0;i<=n/2-1;i++)
	{
		if(a[i]=='?')
		b1++;
		else
		s1+=a[i]-'0';
	}
	for(int i=n/2;i<n;i++)
	{
		if(a[i]=='?')
		b2++;
		else
		s2+=a[i]-'0';
	}
	if(b1==0&&b2==0)
	{
		if(s1==s2)
		cout<<"Bicarp";
		else
		cout<<"Monocarp";
	}
	else if(b1==b2)
	{
		if(fabs(s1-s2)>=1)
		cout<<"Monocarp";
		else
		cout<<"Bicarp";
	}
	else if(b1>b2)
	{
			if(s2-s1==9*(b1-b2)/2)
			cout<<"Bicarp";
			else
			cout<<"Monocarp";
	}
	else if(b1<b2)
	{
		if(s1-s2==9*(b2-b1)/2)
		cout<<"Bicarp";
		else
		cout<<"Monocarp";
	}
	
	
}