A substring of a string T is defined as:

T(i, k)=T_iT_i+1...T_i+k-1, 1≤i≤i+k-1≤|T|.

Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):

S = {(i, j, k) | k≥K, A(i, k)=B(j, k)}.

You are to give the value of |S| for specific A, B and K.

The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.

1 ≤ |A|, |B| ≤ 10⁵
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.

For each case, output an integer |S|.

/** POJ 3415 Common Substrings* 给定两个字符串A和B，求长度不小于k的公共子串的个数* 基本思路是计算A的所有后缀和B的所有后缀之间的最长公共前缀的长度，* 把最长公共前缀长度不小于k的部分全部加起来。* 先把两个字符串连起来，中间用一个没有用过的字符隔开。* 按height分组后，接下来便是快速的统计每组中后缀之间的最长公共前缀之和* 用一个单调的栈来维护，每遇到一个B的后缀就统计与前面的A的后缀* 能产生多少个长度不小于ｋ的公共子串。最A也一样做一边*/
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN=200010;int sa[MAXN];
int t1[MAXN],t2[MAXN],c[MAXN];
int rank[MAXN],height[MAXN];void build_sa(int s[],int n,int m)
{int i,j,p,*x=t1,*y=t2;for(i=0;i<m;i++)c[i]=0;for(i=0;i<n;i++)c[x[i]=s[i]]++;for(i=1;i<m;i++)c[i]+=c[i-1];for(i=n-1;i>=0;i--)sa[--c[x[i]]]=i;for(j=1;j<=n;j<<=1){p=0;for(i=n-j;i<n;i++)y[p++]=i;for(i=0;i<n;i++)if(sa[i]>=j)y[p++]=sa[i]-j;for(i=0;i<m;i++)c[i]=0;for(i=0;i<n;i++)c[x[y[i]]]++;for(i=1;i<m;i++)c[i]+=c[i-1];for(i=n-1;i>=0;i--)sa[--c[x[y[i]]]]=y[i];swap(x,y);p=1;x[sa[0]]=0;for(i=1;i<n;i++)x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;if(p>=n)break;m=p;}
}void getHeight(int s[],int n)
{int i,j,k=0;for(i=0;i<=n;i++)rank[sa[i]]=i;for(i=0;i<n;i++){if(k)k--;j=sa[rank[i]-1];while(s[i+k]==s[j+k])k++;height[rank[i]]=k;}
}
char str1[MAXN],str2[MAXN];
int r[MAXN];
int sta[MAXN],stb[MAXN];
int main()
{freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);int k;int n;int len1,len2;while(scanf("%d",&k)==1 && k){scanf("%s%s",&str1,&str2);len1=strlen(str1);len2=strlen(str2);n=len1+len2+1;for(int i=0;i<len1;i++)r[i]=str1[i];r[len1]=1;for(int i=0;i<len2;i++)r[i+len1+1]=str2[i];r[len1+len2+1]=0;build_sa(r,n+1,128);getHeight(r,n);long long ans=0;long long ss=0;int top=0;for(int i=2;i<=n;i++){if(height[i]<k){ss=0;top=0;continue;}int cnt=0;if(sa[i-1]<len1){cnt++;ss+=height[i]-k+1;}while(top>0 && height[i]<=sta[top-1]){top--;ss-=stb[top]*(sta[top]-height[i]);cnt+=stb[top];}sta[top]=height[i];stb[top++]=cnt;if(sa[i]>len1)ans+=ss;}ss=0;top=0;for(int i=2;i<=n;i++){if(height[i]<k){ss=0;top=0;continue;}int cnt=0;if(sa[i-1]>len1){cnt++;ss+=height[i]-k+1;}while(top>0 && height[i]<=sta[top-1]){top--;ss-=stb[top]*(sta[top]-height[i]);cnt+=stb[top];}sta[top]=height[i];stb[top++]=cnt;if(sa[i]<len1)ans+=ss;}printf("%I64d\n",ans);}return 0;
}