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)

// #define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep(i, n) for (int i = 0; i < n; ++i)
#define REP(i, n) for (int i = 0; i < n; ++i)
#define range(i,a,b) ((a)<=(i) && (i)<(b))
#define debug(x) cout << #x << ' ' << '=' << ' ' << (x) << endl;
#define fs first
#define sc second
#define pb push_back
#define eb emplace_back
typedef long long ll;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> T;
typedef vector<ll> vec;
typedef vector<P> pvec;
typedef vector<vector<ll>> vvec;
typedef vector<vector<P>> pvvec;
typedef priority_queue<ll> PQI;
typedef priority_queue<P> PQP;
typedef priority_queue<ll,vector<ll>,greater<ll>> PQIG;
typedef priority_queue<P,vector<P>,greater<P>> PQPG;
const vector<int> dx = {0, -1, 0, 1, 1, 1, -1, -1};
const vector<int> dy = {1, 0, -1, 0, 1, -1, 1, -1};
const int MOD = (1000000007);
// const int MOD = (998244353);
// const int INF = (1 << 30); // 1073741824
const int INF = (1LL << 60); // 1152921504606846976
const double EPS = (1 >> 30);
template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b; return 1;} return 0;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b; return 1;} return 0;}
template< typename T1, typename T2 > istream &operator>>(istream &is, pair< T1, T2 > &p) { is >> p.first >> p.second; return is; }

struct UnionFind {
  private:
  vector<int> child,tree;
  vector<vector<int>> list;

public:
  UnionFind(int v) {
    tree.resize(v);
    list.resize(v);
    REP(i,v) tree[i] = i, list[i].push_back(i);
  }

  int root(int i){
    if(i == tree[i]){
      REP(j,child.size()) tree[child[j]] = i;
      child.clear(); 
      return i;
    }
    else{
      child.push_back(i);
      return i = root(tree[i]);
    }
  }

  int size(int i){
    return list[root(i)].size();
  }

  vector<int> nodes(int no){
    return list[root(no)];
  }

  bool unit(int x, int y){
    x = root(x);
    y = root(y);
    if(x==y) return false;
    if(list[x].size() < list[y].size()) swap(x,y);
    for(int no:list[y]) list[x].emplace_back(no);
    tree[y] = x;
    return true;
  }

  bool isUnit(int x, int y){
    return root(x) == root(y);
  }
};

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(15);
    //---------------------------------------------
    
    int n;
    cin>>n;
    vector<P> rl(n);
    map<int,int> mp; // 座圧用
    rep(i,n){
        cin>>rl[i].fs>>rl[i].sc;
        mp[rl[i].fs]=1;
        mp[rl[i].sc]=1;
    }
    int idx=0;
    for(auto x:mp){
        mp[x.fs]=idx;
        idx++;
    }
    rep(i,n){
        rl[i].fs=mp[rl[i].fs];
        rl[i].sc=mp[rl[i].sc];
    }
    sort(all(rl)); // 始点でソートしておく
    vector<int> dp(2*n+100);
    dp[0]=1;
    idx=0;
    for(int i=0;i<2*n+10;i++){
        while(rl[idx].fs==i){ // iが始点の区間
            dp[rl[idx].sc+1]+=dp[rl[idx].fs];
            idx++;
        }
        dp[i+1]+=dp[i];
    }
    cout<<dp[rl[n-1].sc+1]<<endl;
}