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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define REP(i,n) for (int i = 1; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define vout(x) rep(i,x.size()) cout << x[i] << " "
template<class T> bool chmin(T &a, T b) {if(a>b) {a=b;return 1;}return 0;}
template<class T> bool chmax(T &a, T b) {if(a<b) {a=b;return 1;}return 0;}
using namespace std;
using vint = vector<int>;
using vvint = vector<vector<int>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<int,int>;
const int inf = 1e9;
const ll inf_l = 1e18;
const int MAX = 1e5;

const int mod = 1e9+7;
struct mint {
    ll x;
    mint (ll x = 0) : x((x%mod+mod)%mod) {}
    mint operator -() const {return mint(-x);}
    mint& operator +=(const mint a) {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator -=(const mint a) {
        if ((x += mod-a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator *=(const mint a) {(x *= a.x) %= mod; return *this;}
    mint operator +(const mint a) const {return mint(*this) += a;}
    mint operator -(const mint a) const {return mint(*this) -= a;}
    mint operator *(const mint a) const {return mint(*this) *= a;}
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    mint inv() const {return pow(mod-2);}
    mint operator /=(const mint a) {return *this *= a.inv();}
    mint operator /(const mint a) {return mint(*this) /= a;}
};
istream& operator >>(istream &is, const mint &a) {return is >> a.x;}
ostream& operator <<(ostream &os, const mint &a) {return os << a.x;}

mint dp[200005];

int main() {
    int n; cin >> n;
    vvint data(200001);
    vint c(n);
    rep(i,n) cin >> c[i];
    c.erase(unique(all(c)),c.end());
    n = c.size();
    rep(i,n) data[c[i]].push_back(i);
    dp[0] = 1;
    REP(i,n+1) {
        dp[i] = dp[i-1];
        int id = lower_bound(all(data[c[i-1]]),i-1)-data[c[i-1]].begin();
        if (id == 0) continue;
        dp[i] += dp[data[c[i-1]][id-1]+1];
    }
    cout << dp[n] << endl;
}