cf299C Weird Game

Weird Game

Yaroslav, Andrey and Roman can play cubes for hours and hours. But the game is for three, so when Roman doesn't show up, Yaroslav and Andrey play another game.

Roman leaves a word for each of them. Each word consists of 2·n binary characters "0" or "1". After that the players start moving in turns. Yaroslav moves first. During a move, a player must choose an integer from 1 to 2·n, which hasn't been chosen by anybody up to that moment. Then the player takes a piece of paper and writes out the corresponding character from his string.

Let's represent Yaroslav's word as s = s₁s₂... s_2n. Similarly, let's represent Andrey's word as t = t₁t₂... t_2n. Then, if Yaroslav choose number k during his move, then he is going to write out character s_k on the piece of paper. Similarly, if Andrey choose number r during his move, then he is going to write out character t_r on the piece of paper.

The game finishes when no player can make a move. After the game is over, Yaroslav makes some integer from the characters written on his piece of paper (Yaroslav can arrange these characters as he wants). Andrey does the same. The resulting numbers can contain leading zeroes. The person with the largest number wins. If the numbers are equal, the game ends with a draw.

You are given two strings s and t. Determine the outcome of the game provided that Yaroslav and Andrey play optimally well.

Input

The first line contains integer n (1 ≤ n ≤ 10⁶). The second line contains string s — Yaroslav's word. The third line contains string t — Andrey's word.

It is guaranteed that both words consist of 2·n characters "0" and "1".

Output

Print "First", if both players play optimally well and Yaroslav wins. If Andrey wins, print "Second" and if the game ends with a draw, print "Draw". Print the words without the quotes.

Example

Input

2
0111
0001

Output

First

Input

3
110110
001001

Output

First

Input

3
111000
000111

Output

Draw

Input

4
01010110
00101101

Output

First

Input

4
01100000
10010011

Output

Second

很显然的贪心策略，
记每个点为（自己能取1，别人能取1），优先取(1,1)，然后有(1,0)优先取，再然后有(0,1)优先取，然后没有然后了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
#define mod 100007
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char s[2000010],t[2000010];
int a,b1,b2,c;
int n,s1,s2;
inline void choose(int &b1,int &b2,int &s1)
{
    if (a){a--;s1++;return;}
    if (b1){b1--;s1++;return;}
    if (b2){b2--;return;}
    c--;
}
int main()
{
    n=read()*2;
    scanf("%s",s+1);
    scanf("%s",t+1);
    for(int i=1;i<=n;i++)
        if (s[i]=='1'&&t[i]=='1')a++;
        else if (s[i]=='1'&&t[i]=='0')b1++;
        else if (s[i]=='0'&&t[i]=='1')b2++;
        else c++;
    n/=2;
    for (int i=1;i<=n;i++)
    {
        choose(b1,b2,s1);
        choose(b2,b1,s2);
    }
    if (s1>s2)puts("First");
    else if (s1==s2)puts("Draw");
    else puts("Second");
}