/*

统计串的出现次数显然可以在自动机上匹配出来即可

但是每次都挨个匹配的话会时间爆炸

那么考虑我们把串复制一份， 然后一起在后缀自动机上跑， 当我们匹配长度大于该串长度的时候强行失配即可

可能会有旋转后相同的串所以开个数组判重 

*/

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<queue>

#include<iostream>

#define ll long long

#define M 2000100

#define mmp make_pair

using namespace std;

int read()

{

	int nm = 0, f = 1;

	char c = getchar();

	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';

	return nm * f;

}

char s[M];

int ch[M][26], fa[M], len[M], sz[M], tot[M], a[M], lst = 1, cnt = 1;

int vis[M];

void insert(int c)

{

	int p = ++cnt, f = lst;

	lst = p;

	len[p] = len[f] + 1;

	sz[p] = 1;

	while(f && !ch[f][c]) ch[f][c] = p, f = fa[f];

	if(!f)

	{

		fa[p] = 1;

	}

	else

	{

		int q = ch[f][c];

		if(len[q] == len[f] + 1)

		{

			fa[p] = q;

		}

		else

		{

			int nq = ++cnt;

			memcpy(ch[nq], ch[q], sizeof(ch[q]));

			fa[nq] = fa[q];

			len[nq] = len[f] + 1;

			fa[p] = fa[q] = nq;

			while(f && ch[f][c] == q) ch[f][c] = nq, f = fa[f];

		}

	}

}

int main()

{

	scanf("%s", s + 1);

	int l = strlen(s + 1);

	for(int i = 1; i <= l; i++) insert(s[i] - 'a');

	for(int i = 1; i <= cnt; i++) tot[len[i]]++;

	for(int i = 1; i <= cnt; i++) tot[i] += tot[i - 1];

	for(int i = 1; i <= cnt; i++) a[tot[len[i]]--] = i;

	for(int i = cnt; i >= 1; i--) sz[fa[a[i]]] += sz[a[i]];

	int m = read();

	for(int cor = 1; cor <= m; cor++)

	{

		scanf("%s", s + 1);

		l = strlen(s + 1);

		for(int i = 1; i <= l; i++) s[l + i] = s[i];

		int now = 1, lenn = 0;

		ll ans = 0;

		for(int i = 1; i <= l * 2; i++)

		{

			int c = s[i] - 'a';

			while(now && !ch[now][c]) now = fa[now], lenn = len[now];

			if(!now) now = 1, lenn = 0;

			else now = ch[now][c], lenn++;

			while(now && len[fa[now]] >= l) now = fa[now], lenn = len[now]; //只能统计那些刚刚超过的节点

			if(lenn >= l && vis[now] != cor) vis[now] = cor, ans += sz[now];

		}

		cout << ans << "\n";

	}

	return 0;

}