Educational Codeforces Round 24 D. Multicolored Cars

D. Multicolored Cars

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice and Bob got very bored during a long car trip so they decided to play a game. From the window they can see cars of different colors running past them. Cars are going one after another.

The game rules are like this. Firstly Alice chooses some color A, then Bob chooses some color B (A ≠ B). After each car they update the number of cars of their chosen color that have run past them. Let's define this numbers after i-th car cntA(i) and cntB(i).

• If cntA(i) > cntB(i) for every i then the winner is Alice.
• If cntB(i) ≥ cntA(i) for every i then the winner is Bob.
• Otherwise it's a draw.

Bob knows all the colors of cars that they will encounter and order of their appearance. Alice have already chosen her color A and Bob now wants to choose such color B that he will win the game (draw is not a win). Help him find this color.

If there are multiple solutions, print any of them. If there is no such color then print -1.

Input

The first line contains two integer numbers n and A (1 ≤ n ≤ 105, 1 ≤ A ≤ 106) – number of cars and the color chosen by Alice.

The second line contains n integer numbers c1, c2, ..., cn (1 ≤ ci ≤ 106) — colors of the cars that Alice and Bob will encounter in the order of their appearance.

Output

Output such color B (1 ≤ B ≤ 106) that if Bob chooses it then he will win the game. If there are multiple solutions, print any of them. If there is no such color then print -1.

It is guaranteed that if there exists any solution then there exists solution with (1 ≤ B ≤ 106).

Examples

input

4 1
2 1 4 2

output

2

input

5 2
2 2 4 5 3

output

-1

input

3 10
1 2 3

output

4

Note

Let's consider availability of colors in the first example:

• cnt2(i) ≥ cnt1(i) for every i, and color 2 can be the answer.
• cnt4(2) < cnt1(2), so color 4 isn't the winning one for Bob.
• All the other colors also have cntj(2) < cnt1(2), thus they are not available.

In the third example every color is acceptable except for 10.

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<cmath>
#include<set>
#include<stack>
#define ll long long
#define pb push_back
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define cls(name,x) memset(name,x,sizeof(name))
using namespace std;
const int inf=1e9+10;
const int maxn=1e6+10;
const int maxm=20;
const int mod=1e9+7;
const double pi=acos(-1.0);
int n,A;
struct node
{
    int id;
    int c;
}color[maxn];
struct cmp
{
    bool operator()(const node &a,const node &b)
    {
            return a.c<b.c;
    }
};
int ans[maxn];
int del[maxn];
int main()
{
    //freopen("in.txt","r",stdin);
    while(~scanf("%d %d",&n,&A))
    {
        multiset<node,cmp> S;
        for(int i=1;i<=1e6;i++)
        {
            color[i].id=i;
            color[i].c=0;
            S.insert(color[i]);
        }
        for(int i=1;i<=1e6;i++)
            ans[i]=1;
        ans[A]=0;
        multiset<node,cmp>::iterator it,ttt;
        for(int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&x);
            if(x==A||ans[x]==1)
            {
                S.erase(color[x]);
                color[x].c++;
                S.insert(color[x]);
            }
            int k=0;
            for(it=S.begin();(*it).c<color[A].c;it++)
                del[k++]=(*it).id;
            for(int j=0;j<k;j++)
            {
                ans[del[j]]=0;
                S.erase(color[del[j]]);
            }
        }
        int flag=0;
        for(int i=1;i<=1e6;i++)
            if(ans[i]!=0) flag=i;
        if(flag==0) printf("-1");
        else printf("%d
",flag);
    }
    return 0;
}