CodeForces 163A

One day Polycarpus got hold of two non-empty strings s and t, consisting of lowercase Latin letters. Polycarpus is quite good with strings, so he immediately wondered, how many different pairs of "x y" are there, such that x is a substring of string s, y is a subsequence of string t, and the content of x and y is the same. Two pairs are considered different, if they contain different substrings of string sor different subsequences of string t. Read the whole statement to understand the definition of different substrings and subsequences.

The length of string s is the number of characters in it. If we denote the length of the string s as |s|, we can write the string as s = s₁s₂... s_|s|.

A substring of s is a non-empty string x = s[a... b] = s_as_a + 1... s_b (1 ≤ a ≤ b ≤ |s|). For example, "code" and "force" are substrings or "codeforces", while "coders" is not. Two substrings s[a... b] and s[c... d] are considered to be different if a ≠ cor b ≠ d. For example, if s="codeforces", s[2...2] and s[6...6] are different, though their content is the same.

A subsequence of s is a non-empty string y = s[p₁p₂... p_|y|] = s_p1s_p2... s_p|y| (1 ≤ p₁ < p₂ < ... < p_|y| ≤ |s|). For example, "coders" is a subsequence of "codeforces". Two subsequences u = s[p₁p₂... p_|u|] and v = s[q₁q₂... q_|v|] are considered different if the sequences p and q are different.

Input

The input consists of two lines. The first of them contains s (1 ≤ |s| ≤ 5000), and the second one contains t (1 ≤ |t| ≤ 5000). Both strings consist of lowercase Latin letters.

Output

Print a single number — the number of different pairs "x y" such that x is a substring of string s, y is a subsequence of string t, and the content of x and y is the same. As the answer can be rather large, print it modulo 1000000007 (10⁹ + 7).

Example

Input

aaaa

Output

5

Input

codeforcesforceofcode

Output

60

Note

Let's write down all pairs "x y" that form the answer in the first sample: "s[1...1]t[1]", "s[2...2] t[1]", "s[1...1] t[2]","s[2...2] t[2]", "s[1...2] t[1 2]".

题目大意：两个字符串，求第一个的所有子串（连续）与第二个的所有子序列（不一定连续）相同的情况之和；如果用最容易想到的办法：先枚举一个串的每一个子串，用枚举的这个子串与另一字符串的子串匹配，依次进行下去，肯定会超时。

看到第一眼会想到和LCS（最长公共子序列）很像，只不过要求第一个字符串必须在原字符串中连续，LCS是求最长的匹配字符串，本题就转化为求最大匹配数.

状态方程：dp[i][j]表示2号串中以i结尾的子序列与1号串中以j结尾的子串相等的所有情况数目；

状态转移方程：两种可转移状态：一：str1[j]!=str2[i]，因为没有匹配上，dp[i][j]也就可以表示为2号串中以i-1结尾的子序列与1号串中以j结尾的子串相匹配的状况；即：dp[i][j]=dp[i-1][j];

二：我们把i结尾的子序列与j结尾的子串未匹配之前的状态也可以置为dp[i][j]=dp[i-1][j]当str1[j]=str2[i]时：情况的数目会增加：dp[i-1][j-1]+1;dp[i][j]=未匹配前状态下的情况数目+匹配成功增加的情况数目。

那么结果就等于，对每次匹配成功后新增加的情况数求和；

ans+=dp[i-1][j-1]+1;

#include<bits/stdc++.h>using namespace std;#define N 5005#define mod 1000000007typedef long long ll;char a[N],b[N];int lena,lenb;ll dp[N][N];int main(){    int i,j;    while(~scanf("%s%s",b+1,a+1))    {        lena=strlen(a+1);        lenb=strlen(b+1);        ll ans=0;        for(i=1;i<=lena;i++)        {            for(j=1;j<=lenb;j++)            {                 dp[i][j]=dp[i-1][j];               if(a[i]==b[j])                  {                     // cout<<dp[i][j]<<"---"<<dp[i-1][j-1]+1<<" "<<i<<"--"<<j<<endl;                      dp[i][j]=(dp[i][j]+dp[i-1][j-1]+1)%mod;                      ans=(ans+dp[i-1][j-1]+1)%mod;                  }            }        }        printf("%I64d\n",ans);    }    return 0;}