Now you have a string consists of uppercase letters, two integers AA and BB. We call a substring wonderful substring when the times it appears in that string is between AA and BB (A \le times \le BA≤times≤B). Can you calculate the number of wonderful substrings in that string?

Input

Input has multiple test cases.

For each line, there is a string SS, two integers AA and BB.

\sum length(S) \le 2 \times 10^6∑length(S)≤2×106,

1 \le A \le B \le length(S)1≤A≤B≤length(S)

Output

For each test case, print the number of the wonderful substrings in a line.

樣例輸入復制

AAA 2 3
ABAB 2 2

樣例輸出復制

2
3

題目來源

ACM-ICPC 2018 焦作賽區網絡預賽

題解：SAM模板題

參考代碼：

//H  求子串出現次數在k1=<num<=k2;
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4e5+10;
char ss[200005];
const int LetterSize = 26;

int tot, last,ch[MAXN][LetterSize],fa[MAXN],len[MAXN];
int sum[MAXN],tp[MAXN],cnt[MAXN]; 

void init()
{
    last = tot = 1;
    len[1] = 0;
    memset(ch,0,sizeof ch);
    memset(fa,0,sizeof fa);
    memset(cnt,0,sizeof cnt);
}

void add( int x)
{
    int p = last, np = last = ++tot;
    len[np] = len[p] + 1, cnt[last] = 1;
    while( p && !ch[p][x]) ch[p][x] = np, p = fa[p];
    if(p == 0) fa[np] = 1;
    else
    {
        int q = ch[p][x];
        if( len[q] == len[p] + 1)
            fa[np] = q;
        else
        {
            int nq = ++tot;
            memcpy( ch[nq], ch[q], sizeof ch[q]);
            len[nq] = len[p] + 1, fa[nq] = fa[q], fa[q] = fa[np] = nq;
            while( p && ch[p][x] == q)  ch[p][x] = nq, p = fa[p];
        }
    }
}

void toposort()
{
    for(int i = 1; i <= len[last]; i++)   sum[i] = 0;
    for(int i = 1; i <= tot; i++)   sum[len[i]]++;
    for(int i = 1; i <= len[last]; i++)   sum[i] += sum[i-1];
    for(int i = 1; i <= tot; i++)   tp[sum[len[i]]--] = i;
}


int main()
{

    int k1,k2;
	while(scanf("%s",ss)!=EOF)
    {
    	init();
    	scanf("%d%d",&k1,&k2);
	    long long ans=0;
        for(int i=0,len=strlen(ss);i<len;i++) add(ss[i]-‘A‘);
        toposort();
        for(int i=tot;i;i--)
        {
            int p=tp[i],fp=fa[p];
            cnt[fp]+=cnt[p];
            if(cnt[p]>=k1 && cnt[p]<=k2) ans+=len[p]-len[fp];
        }
        printf("%lld\n",ans);
    }

    return 0;
}

ACM-ICPC 2018 焦作賽區網絡預賽 H題 String and Times（SAM）