Arthur's Language

CF-101628-A

Arthur is an extremely important person and rarely has the time to talk. To minimize the time spent with words, he developed his own language, which he uses on a daily basis. However, it’s very hard to understand whatever comes out of his mouth.

In order to improve his interaction with the rest of the world, Arthur has asked the Narratives Industrial Complex (CIN in Portuguese) to develop a program which receives a speech S produced by Arthur and a pattern w and prints how many distinct ways we can obtain w by only removing characters from S. Two ways are considered distinct if there is an index 0 ≤ i < |S| such as Si is present in one and not in the other.

Since you are considered the most experienced programmer in CIN, your team has trusted you the task of writing this program.

Input

The first line consists of a string S(1 ≤ |S| ≤ 105), representing Arthur’s speech. S contains only lower and upper case letters.
The second and last line consists of a string w(1 ≤ |w| ≤ 10), the pattern to be searched.

Output

Print a single integer: The number of distinct ways to obtain w by removing letters from S. Because this number can be large, print the remained of this number divided by 1000000007 = 109 + 7

Examples

Input

weewrshkim
sim

Output

1

Input

qqqaaabbbcccfffrrr
qabcfr

Output

729

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题意

有两个字符串：a，b
在a中移除一些字符可得到b
问：有多少中方案

---

题解

因为是移除一些字符，所以原串的顺序是不变的
dp递推：
dp[i][j]表示模式串中第i个字符匹配到匹配串中第j个字符时有多少方案数
故dp[a.size()-1]dp[b.size()-1]即为所求

#include<iostream>
#include<string>
using namespace std;
typedef long long ll;
ll dp[20][100000 + 50];
const ll MOD = 1e9 + 7;
int main()
{
    string s, p;
    cin >> s >> p;
    for (int i = 0; i < s.size(); i++)//预处理第一行
    {
        if (i)
            dp[0][i] = dp[0][i - 1];
        if (s[i] == p[0])
        {
            dp[0][i]++;
        }
    }//是一个前缀和
    for (int i = 1; i < p.size(); i++)
    {
        for (int j = i; j < s.size(); j++)
        {
            if (j)
                dp[i][j] = dp[i][j - 1];//维护前缀和
            if (p[i] == s[j])
            {
                dp[i][j] += dp[i - 1][j - 1];//不能是dp[i-1][j]
                dp[i][j] %= MOD;
            }
        }
    }
    cout << dp[p.size() - 1][s.size() - 1] % MOD << endl;
}