#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//ll inv(ll a, ll p) {
//	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
//}
//modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };

void Z_algorithm(const string &s, vector<int> &a) {
	int sz = s.size();
	a.resize(sz);
	a[0] = sz;
	int i = 1, j = 0;
	while (i < sz) {
		while (i + j < sz&&s[j] == s[i + j])++j;
		a[i] = j;
		if (j == 0) { ++i; continue; }
		int k = 1;
		while (i + k < sz&&k + a[k] < j)a[i + k] = a[k], ++k;
		i += k; j -= k;
	}
}

struct Node {
	Node *ch[26];
	int las;
	Node() {
		rep(i, 26)ch[i] = NULL; las = 0;
	}
};

void add(Node *nd, string &s) {
	rep(i, s.size()) {
		int t = s[i] - 'a';
		if (nd->ch[t] == NULL)nd->ch[t] = new Node();
		nd = nd -> ch[t];
	}
	nd->las = 1;
}

const int b = 500;
modint dp[1 << 17];
vector<int> r[1 << 17];
void solve() {
	int n; cin >> n;
	vector<string> v(n);
	Node* root = new Node();
	vector<string> ad;
	rep(i, n) {
		cin >> v[i];
		if (v[i].size() <= b) {
			add(root, v[i]);
		}
		else {
			ad.push_back(v[i]);
		}
	}
	string s; cin >> s;
	n = s.size();
	rep(i, ad.size()) {
		string t = ad[i] + s;
		vector<int> ra;
		Z_algorithm(t, ra);
		rep(j, n) {
			if (ra[j + ad[i].size()] >= ad[i].size()) {
				r[j].push_back(ad[i].size());
			}
		}
	}
	dp[0] = 1;
	rep(i, n) {
		for (int len : r[i]) {
			dp[i + len] += dp[i];
		}
		Node *cur = root;
		rep(j, b) {
			if (i + j >= n)break;
			int nex = s[i + j] - 'a';
			if (cur->ch[nex] == NULL)break;
			cur = cur->ch[nex];
			if (cur->las)dp[i + j + 1] += dp[i];
		}
	}
	cout << dp[n] << endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(7);
	//init();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}


### Prompt

In CPP, your task is to solve 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<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = (1e+18) + 7;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-6;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%mod;
		a = a * a%mod; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

//ll inv(ll a, ll p) {
//	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
//}
//modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

int dx[4] = { 1,0,-1,0 };
int dy[4] = { 0,1,0,-1 };

void Z_algorithm(const string &s, vector<int> &a) {
	int sz = s.size();
	a.resize(sz);
	a[0] = sz;
	int i = 1, j = 0;
	while (i < sz) {
		while (i + j < sz&&s[j] == s[i + j])++j;
		a[i] = j;
		if (j == 0) { ++i; continue; }
		int k = 1;
		while (i + k < sz&&k + a[k] < j)a[i + k] = a[k], ++k;
		i += k; j -= k;
	}
}

struct Node {
	Node *ch[26];
	int las;
	Node() {
		rep(i, 26)ch[i] = NULL; las = 0;
	}
};

void add(Node *nd, string &s) {
	rep(i, s.size()) {
		int t = s[i] - 'a';
		if (nd->ch[t] == NULL)nd->ch[t] = new Node();
		nd = nd -> ch[t];
	}
	nd->las = 1;
}

const int b = 500;
modint dp[1 << 17];
vector<int> r[1 << 17];
void solve() {
	int n; cin >> n;
	vector<string> v(n);
	Node* root = new Node();
	vector<string> ad;
	rep(i, n) {
		cin >> v[i];
		if (v[i].size() <= b) {
			add(root, v[i]);
		}
		else {
			ad.push_back(v[i]);
		}
	}
	string s; cin >> s;
	n = s.size();
	rep(i, ad.size()) {
		string t = ad[i] + s;
		vector<int> ra;
		Z_algorithm(t, ra);
		rep(j, n) {
			if (ra[j + ad[i].size()] >= ad[i].size()) {
				r[j].push_back(ad[i].size());
			}
		}
	}
	dp[0] = 1;
	rep(i, n) {
		for (int len : r[i]) {
			dp[i + len] += dp[i];
		}
		Node *cur = root;
		rep(j, b) {
			if (i + j >= n)break;
			int nex = s[i + j] - 'a';
			if (cur->ch[nex] == NULL)break;
			cur = cur->ch[nex];
			if (cur->las)dp[i + j + 1] += dp[i];
		}
	}
	cout << dp[n] << endl;
}

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(7);
	//init();
	//int t; cin >> t; rep(i, t)solve();
	solve();
	stop
		return 0;
}


```