Problem:
Score : 700 points

Problem Statement

There are N stones arranged in a row. The i-th stone from the left is painted in the color C_i.

Snuke will perform the following operation zero or more times:

Choose two stones painted in the same color. Repaint all the stones between them, with the color of the chosen stones.

Find the number of possible final sequences of colors of the stones, modulo 10^9+7.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq C_i \leq 2\times 10^5(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1
:
C_N

Output

Print the number of possible final sequences of colors of the stones, modulo 10^9+7.

Sample Input 1

5
1
2
1
2
2

Sample Output 1

3

We can make three sequences of colors of stones, as follows:

(1,2,1,2,2), by doing nothing.

(1,1,1,2,2), by choosing the first and third stones to perform the operation.

(1,2,2,2,2), by choosing the second and fourth stones to perform the operation.

Sample Input 2

6
4
2
5
4
2
4

Sample Output 2

5

Sample Input 3

7
1
3
1
2
3
3
2

Sample Output 3

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//#include "pch.h"
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

#include <vector>
#include <set>
#include <map>
#include <iostream>
#include <utility>
#include <queue>
#include <algorithm>
#include <functional>
#include <string>
#include <bitset>

#define ll long long
#define fri(n) for(i=0;i<(n);i++)
#define frj(n) for(j=0;j<(n);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000000000//10^18
#define INFINT 2000000001//2*10^9+1
#define MOD 1000000007
#define MODANOTHER 998244353
#define PI acos(-1)

using namespace std;

static ll dp[200010];
static vector<int> numwhere[200010];
static int c[200010];
int n;



ll search(int x){
	if(x==n-1){
		dp[x]=1;
		return 1;
	}
	if(dp[x]!=0) return dp[x];

	ll ret;
	int temp;
	int ng,ok;

	ng=numwhere[c[x]].size();
	ok=0;

	while(ng-ok>1){
		temp=(ok+ng)/2;
		if(numwhere[c[x]][temp]<=x){
			ok=temp;
		}else{
			ng=temp;
		}
	}

	if(numwhere[c[x]].size()>ok+1){
		ret=search(x+1)+search(numwhere[c[x]][ok+1]);
	}else{
		ret=search(x+1);
	}

	dp[x]=ret;

	return ret;
}

int main(void)
{
	//変数の宣言
//	int n;
//	static int c[200010];
//	static vector<int> numwhere[200010];

	//よく使う変数
	int i,j,k,l;
	int flag=0;
	int ans=0;
	int count=0;
	int temp,temp1,temp2;
	int max,min;
	int len;
	int sum=0;
	int ok,ng;
	int r;
	


	//データの読み込み


	scanf("%d",&n);
//	scanf_s("%d",&n);

	for(i=0;i<n;i++){
		scanf("%d",&c[i]);
//		scanf_s("%d",&c[i]);
	}
	
	//	printf("nは%dです\n", n);
	//	printf("データの読み込み終了\n");
	//実際の処理
	
	temp=0;
	for(i=1;i<n;i++){
		if(c[i-1]!=c[i]){
			c[i-temp]=c[i];
		}else{
			temp++;
		}
	}

	n=n-temp;

	for(i=0;i<n;i++){
		numwhere[c[i]].push_back(i);
	}

	search(0);

	//	printf("計算部分終了\n");

	//出力

/*	for(i=0;i<n;i++){
		printf("dp[%d]=%lld\n",i,dp[i]);
	}*/

	printf("%lld",dp[0]);


	//	printf("結果の出力終了\n");


	return 0;
}