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: Accepted)

#include <algorithm>
#include <cmath>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;
using LLONG = long long;
const LLONG MOD = 1000000007;

int main()
{
    int N; cin >> N;
    vector<int> cs(N + 1);
    cs[0] = -1;
    for (int i = 1; i <= N; ++i)
    {
        cin >> cs[i];
    }
    // dp[i]:i番目の石まで考えたときの塗り分け方の総和
    vector<LLONG> dp(N + 1, 0);
    // 各色に対するdp[j](j<i)の総和
    map<int, LLONG> color2dpSumMap;
    for (const int ci : cs)
    {
        color2dpSumMap[ci] = 0;
    }
    dp[0] = 1;
    for (int i = 1; i <= N; ++i)
    {
        // i番目を右端とする塗り替えができる場合
        // これまでの塗り分け方に(i-1)番目までの塗り方の総和を足したものが総塗り分け方
        if (cs[i - 1] != cs[i])
        {
            dp[i] = color2dpSumMap[cs[i]] + dp[i - 1];
            color2dpSumMap[cs[i]] += dp[i - 1];
        }
        // 前の石と色が同じ場合は，どちらを右端にしても変わらないので塗り分け方も引き継ぎ
        else
        {
            dp[i] = color2dpSumMap[cs[i]];
        }
        dp[i] %= MOD;
        color2dpSumMap[cs[i]] %= MOD;
        //// i番目を右端とした塗り替えを考える
        //// j番目までを飛ばし，j番目より後を塗り替えられるとき，
        //// （それ以前（j番目まで）の塗り方）×（j番目より後～i番目までの塗り方：１通り）だけ
        //// dp[i]を増やす
        //// ->単純に実装するとO(N^2)で間に合わない！
        //for (int j = 0; j < i; ++j)
        //{
        //    if (cs[j] != cs[i] && cs[j + 1] == cs[i])
        //    {
        //        dp[i] += dp[j];
        //    }
        //}
    }
    cout << color2dpSumMap[cs.back()] << endl;
}