#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl
#define svec(v) cout<<#v<<":";rep(pachico,v.size())cout<<" "<<v[pachico];cout<<endl
#define svecp(v) cout<<#v<<":";each(pachico,v)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(pachico,s)cout<<" "<<pachico;cout<<endl
#define smap(m) cout<<#m<<":";each(pachico,m)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 200005;

struct RollingHash {
	static const ll mo0=1000000007,mo1=1000000009; static ll mul0,mul1;
	static const ll add0=1000010007,add1=1003333331; static vector<ll> pmo[2];
	string s; int sz; vector<ll> hash_[2];
	void init(string s) {
		this->s=s; sz=(int)s.size();
		hash_[0].resize(sz+1,0),hash_[1].resize(sz+1,0);
		if(!mul0) mul0=10009+(((ll)&mul0)>>5)%259,mul1=10007+(((ll)&mul1)>>5)%257;
		if(pmo[0].empty()) pmo[0].pb(1),pmo[1].pb(1);
		rep(i,sz) hash_[0][i+1]=(hash_[0][i]*mul0+add0+s[i])%mo0;	//hash_[0][i]はインデックス0~i-1までの文字列のハッシュ値
		rep(i,sz) hash_[1][i+1]=(hash_[1][i]*mul1+add1+s[i])%mo1;
	}
	pair<ll,ll> hash(int l,int r) {	//文字列sのインデックス[l,r]の部分文字列のハッシュ値
		if(l>r) return make_pair(0,0);
		while((int)pmo[0].size()<r+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
		return make_pair((hash_[0][r+1]+(mo0-hash_[0][l]*pmo[0][r+1-l]%mo0))%mo0,
			             (hash_[1][r+1]+(mo1-hash_[1][l]*pmo[1][r+1-l]%mo1))%mo1);
	}
	pair<ll,ll> hash(string s) { init(s); return hash(0,(int)s.size()-1); }	//文字列s全体のハッシュ値
	static pair<ll,ll> concat(pair<ll,ll> L,pair<ll,ll> R,int RL) { //文字列L+Rのハッシュ値,RLはRの文字列の長さ
		while((int)pmo[0].size()<RL+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
		return make_pair((R.first + L.first*pmo[0][RL])%mo0,(R.second + L.second*pmo[1][RL])%mo1);
	}
};
vector<ll> RollingHash::pmo[2]; ll RollingHash::mul0,RollingHash::mul1;

set<pll> st[MAX_N];
ll dp[MAX_N/2];

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> l;
    rep(i,n){
        string s;
        cin >> s;
        RollingHash rh;
        st[len(s)].insert(rh.hash(s));
        l.pb(len(s));
    }
    zip(l);
    string t;
    cin >> t;
    int m = len(t);
    RollingHash rh;
    rh.init(t);
    dp[0] = 1;
    rep(i,m){
        if(dp[i]){
            each(j,l){
                if(i+j > m) break;
                if(st[j].find(rh.hash(i,i+j-1)) != st[j].end()){
                    dp[i+j] = (dp[i+j]+dp[i]) % MOD;
                }
            }
        }
    }
    cout << dp[m] << "\n";
    return 0;
}


### Prompt

Please create a solution in Cpp to the following problem:

You are given a string $t$ and a set $S$ of $N$ different strings. You need to separate $t$ such that each part is included in $S$.

For example, the following 4 separation methods satisfy the condition when $t = abab$ and $S = \\{a, ab, b\\}$.

* $a,b,a,b$
* $a,b,ab$
* $ab,a,b$
* $ab,ab$



Your task is to count the number of ways to separate $t$. Because the result can be large, you should output the remainder divided by $1,000,000,007$.



Input

The input consists of a single test case formatted as follows.


$N$
$s_1$
:
$s_N$
$t$


The first line consists of an integer $N$ ($1 \leq N \leq 100,000$) which is the number of the elements of $S$. The following $N$ lines consist of $N$ distinct strings separated by line breaks. The $i$-th string $s_i$ represents the $i$-th element of $S$. $s_i$ consists of lowercase letters and the length is between $1$ and $100,000$, inclusive. The summation of length of $s_i$ ($1 \leq i \leq N$) is at most $200,000$. The next line consists of a string $t$ which consists of lowercase letters and represents the string to be separated and the length is between $1$ and $100,000$, inclusive.

Output

Calculate the number of ways to separate $t$ and print the remainder divided by $1,000,000,007$.

Examples

Input

3
a
b
ab
abab


Output

4


Input

3
a
b
c
xyz


Output

0


Input

7
abc
ab
bc
a
b
c
aa
aaabcbccababbc


Output

160


Input

10
a
aa
aaa
aaaa
aaaaa
aaaaaa
aaaaaaa
aaaaaaaa
aaaaaaaaa
aaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa


Output

461695029

### Response

```cpp
#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto& (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define sar(a,n) cout<<#a<<":";rep(pachico,n)cout<<" "<<a[pachico];cout<<endl
#define svec(v) cout<<#v<<":";rep(pachico,v.size())cout<<" "<<v[pachico];cout<<endl
#define svecp(v) cout<<#v<<":";each(pachico,v)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl
#define sset(s) cout<<#s<<":";each(pachico,s)cout<<" "<<pachico;cout<<endl
#define smap(m) cout<<#m<<":";each(pachico,m)cout<<" {"<<pachico.first<<":"<<pachico.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 200005;

struct RollingHash {
	static const ll mo0=1000000007,mo1=1000000009; static ll mul0,mul1;
	static const ll add0=1000010007,add1=1003333331; static vector<ll> pmo[2];
	string s; int sz; vector<ll> hash_[2];
	void init(string s) {
		this->s=s; sz=(int)s.size();
		hash_[0].resize(sz+1,0),hash_[1].resize(sz+1,0);
		if(!mul0) mul0=10009+(((ll)&mul0)>>5)%259,mul1=10007+(((ll)&mul1)>>5)%257;
		if(pmo[0].empty()) pmo[0].pb(1),pmo[1].pb(1);
		rep(i,sz) hash_[0][i+1]=(hash_[0][i]*mul0+add0+s[i])%mo0;	//hash_[0][i]はインデックス0~i-1までの文字列のハッシュ値
		rep(i,sz) hash_[1][i+1]=(hash_[1][i]*mul1+add1+s[i])%mo1;
	}
	pair<ll,ll> hash(int l,int r) {	//文字列sのインデックス[l,r]の部分文字列のハッシュ値
		if(l>r) return make_pair(0,0);
		while((int)pmo[0].size()<r+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
		return make_pair((hash_[0][r+1]+(mo0-hash_[0][l]*pmo[0][r+1-l]%mo0))%mo0,
			             (hash_[1][r+1]+(mo1-hash_[1][l]*pmo[1][r+1-l]%mo1))%mo1);
	}
	pair<ll,ll> hash(string s) { init(s); return hash(0,(int)s.size()-1); }	//文字列s全体のハッシュ値
	static pair<ll,ll> concat(pair<ll,ll> L,pair<ll,ll> R,int RL) { //文字列L+Rのハッシュ値,RLはRの文字列の長さ
		while((int)pmo[0].size()<RL+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
		return make_pair((R.first + L.first*pmo[0][RL])%mo0,(R.second + L.second*pmo[1][RL])%mo1);
	}
};
vector<ll> RollingHash::pmo[2]; ll RollingHash::mul0,RollingHash::mul1;

set<pll> st[MAX_N];
ll dp[MAX_N/2];

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vector<int> l;
    rep(i,n){
        string s;
        cin >> s;
        RollingHash rh;
        st[len(s)].insert(rh.hash(s));
        l.pb(len(s));
    }
    zip(l);
    string t;
    cin >> t;
    int m = len(t);
    RollingHash rh;
    rh.init(t);
    dp[0] = 1;
    rep(i,m){
        if(dp[i]){
            each(j,l){
                if(i+j > m) break;
                if(st[j].find(rh.hash(i,i+j-1)) != st[j].end()){
                    dp[i+j] = (dp[i+j]+dp[i]) % MOD;
                }
            }
        }
    }
    cout << dp[m] << "\n";
    return 0;
}


```