Problem: Whoso in ignorance draws near to them and hears the Sirens' voice, he nevermore returns.

Homer, Odyssey

In the times of Jason and the Argonauts, it was well known that sirens use the sound of their songs to lure sailors into their demise. Yet only a few knew that every time sirens call a sailor by his name, his will weakens, making him more vulnerable. 

For the purpose of this problem, both siren songs and names of the sailors will be represented as strings of lowercase English letters. The more times the sailor's name occurs as a contiguous substring of the song, the greater danger he is in.

Jason found out that sirens can sing one of the n+1 songs, which have the following structure: let s_i (0 ≤ i ≤ n) be the i-th song and t be a string of length n, then for every i < n: s_{i+1} = s_i t_i s_i. In other words i+1-st song is the concatenation of i-th song, i-th letter (0-indexed) of t and the i-th song. 

Fortunately, he also knows s_0 and t. Jason wonders how many times a sailor's name is mentioned in a particular song. Answer q queries: given the sailor's name (w) and the index of a song (i) output the number of occurrences of w in s_i as a substring. As this number can be quite large, output its remainder modulo 10^9+7.

Input

In the first line of input there are two integers n, q (  1 ≤ n, q ≤ 10^5) meaning that there are n+1 songs and q queries. In the next two lines strings s_0 and t follow (1 ≤ |s_0| ≤ 100, |t| = n). 

Next q lines describe the queries; each one contains an integer k ( 0 ≤ k ≤ n), the index of the song sung by the sirens, and a non-empty string w, which is the name of a sailor. All strings in this problem consist only of lowercase English letters, and the sum of lengths of sailors' names does not exceed 10^6.

Output

Output q lines, i-th of them should contain the remainder modulo 10^9+7 of the number of occurrences of w in s_k.

Examples

Input


3 3
aa
bcd
2 aba
3 ca
3 aa


Output


2
2
8


Input


4 5
aba
bbac
1 a
3 baca
3 ab
2 bab
4 aba


Output


4
0
14
6
28

Note

In the first example songs of the sirens are as follows: 

  * Song 0: aa
  * Song 1: aabaa
  * Song 2: aabaacaabaa
  * Song 3: aabaacaabaadaabaacaabaa

Solution (in CPP):

Solution:
/* by Hyperhydrochloric Acid */
#include <bits/stdc++.h>
using namespace std;
#define SZ(x)               ((int)(x).size())
#define all(x)              (x).begin(), (x).end()
#define loop(i, n)          for(int i = 0; i < (n); ++i)
#define cont(i, n)          for(int i = 1; i <= (n); ++i)
#define circ(i, a, b)       for(int i = (a); i <= (b); ++i)
#define range(i, a, b, c)   for(int i = (a); ((c) > 0 ? i <= (b) : i >= (b)); i += (c))
#define foreach(it, v)      for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
#define y0          y0O0OO00OO0OO0OO0OOO00OO0OO0O0O000OO0
#define y1          y1II11II11III11I1III11II111IIII1II1I1
#define pub         push_back
#define pob         pop_back
#define mak         make_pair
typedef long long   ll;
typedef long double lf;
const int   Inf = 0x3f3f3f3f;
const ll    INF = 0x3f3f3f3f3f3f3f3fll;
/* Source code starts here */

const int qwq = 1 << 20;

template<int jt> struct ModInt {
	int x;
	ModInt(int v = 0) {x = v; if(x >= jt || x < 0) {x %= jt; if(x < 0) x += jt;}}
	ModInt(ll v) {x = v % jt; if(x < 0) x += jt;}
	operator int() const {return x;}
	friend inline ModInt &operator+=(ModInt &a, const ModInt &b) {a.x += b.x; if(a.x >= jt) a.x -= jt; return a;}
	friend inline ModInt operator+(ModInt a, const ModInt &b) {return a += b;}
	friend inline ModInt operator+(ModInt a, int b) {return a += ModInt(b);}
	friend inline ModInt operator+(int a, ModInt b) {return b += ModInt(a);}
	friend inline ModInt &operator++(ModInt &a) {return a += 1;}
	friend inline ModInt &operator-=(ModInt &a, const ModInt &b) {a.x -= b.x; if(a.x < 0) a.x += jt; return a;}
	friend inline ModInt operator-(ModInt a, const ModInt &b) {return a -= b;}
	friend inline ModInt operator-(ModInt a, int b) {return a -= ModInt(b);}
	friend inline ModInt operator-(int a, ModInt b) {return ModInt(a) - b;}
	friend inline ModInt &operator--(ModInt &a) {return a -= 1;}
	friend inline ModInt &operator*=(ModInt &a, const ModInt &b) {a.x = (ll)a.x * b.x % jt; return a;}
	friend inline ModInt operator*(ModInt a, const ModInt &b) {return a *= b;}
	friend inline ModInt operator*(ModInt a, int b) {return a *= ModInt(b);}
	friend inline ModInt operator*(int a, ModInt b) {return b *= ModInt(a);}
	static inline void exgcd(ll a, ll b, ll &x, ll &y) {b ? (exgcd(b, a % b, y, x), y -= a / b * x) : (x = 1, y = 0);}
	friend inline ModInt inv(const ModInt &a) {ll x, y; exgcd(a, jt, x, y); return x;}
	friend inline ModInt operator/(const ModInt &a, const ModInt &b) {return a * inv(b);}
	friend inline ModInt operator/(ModInt a, int b) {return a * inv(ModInt(b));}
	friend inline ModInt operator/(int a, ModInt b) {return ModInt(a) * inv(b);}
	friend inline ModInt operator/=(ModInt &a, const ModInt &b) {return a = a / b;}
};
const int jt = 1000000007;
typedef ModInt<jt> mint;

char tmp[qwq];
void inline gstr(string &s) {
	scanf("%s", tmp);
	s = tmp;
}

template<int ch, int jt> struct strHash {
	vector<int> ha, mii;
	void inline init(const string &s) {
		ha.clear(); mii.clear();
		ha.pub(0); mii.pub(1);
		loop(i, SZ(s)) {
			ha.pub((1ll * ha.back() * ch + s[i]) % jt);
			mii.pub(1ll * mii.back() * ch % jt);
		}
	}
	int inline get(int l, int r) {
		return (ha[r] - 1ll * mii[r - l + 1] * ha[l - 1] % jt + jt) % jt;
	}
};

int n, q, k;
string s, t, w;
vector<string> S;
mint sm[26][1 << 17];
mint mii[1 << 17];
strHash<233, 1000000007> ha1, hw1;
strHash<233, 998244353> ha2, hw2;

int main() {
	scanf("%d%d", &n, &q);
	gstr(s); gstr(t);
	S.pub(s); string tmp = s;
	loop(i, n) {
		tmp = tmp + t[i] + tmp;
		S.pub(tmp);
		if(SZ(tmp) > qwq) break;
	}
	loop(i, 26) loop(j, n) sm[i][j + 1] = sm[i][j] * 2 + (t[j] == i + 'a');
	mii[0] = 1;
	cont(i, n) mii[i] = mii[i - 1] * 2;
	while(q--) {
		scanf("%d", &k);
		gstr(w);
		int pos = 0;
		while(SZ(S[pos]) < SZ(w)) ++pos;
		if(pos > k) {
			puts("0"); continue;
		}
		ha1.init(S[pos]); ha2.init(S[pos]);
		hw1.init(w); hw2.init(w);
		int sw = SZ(w), si = SZ(S[pos]);
		int hh1 = hw1.get(1, sw);
		int hh2 = hw2.get(1, sw);
		mint ot = 0;
		cont(i, SZ(S[pos]) - sw + 1) if(ha1.get(i, i + sw - 1) == hh1 && ha2.get(i, i + sw - 1) == hh2) ++ot;
		ot *= mii[k - pos];
		vector<bool> pre(sw + 2), suf(sw + 2);
		pre[0] = suf[sw + 1] = 1;
		cont(i, sw) {
			if(ha1.get(si - i + 1, si) == hw1.get(1, i) && ha2.get(si - i + 1, si) == hw2.get(1, i)) {
				pre[i] = 1;
			}
			if(ha1.get(1, i) == hw1.get(sw - i + 1, sw) && ha2.get(1, i) == hw2.get(sw - i + 1, sw)) {
				suf[sw - i + 1] = 1;
			}
		}
		cont(i, sw) if(pre[i - 1] && suf[i + 1]) {
			int ch = w[i - 1] - 'a';
			ot += sm[ch][k] - sm[ch][pos] * mii[k - pos];
		}
		printf("%d\n", (int)ot);
	}
	return 0;
}