Chef has a nice complete binary tree in his garden. Complete means that each node has exactly twosons, so the tree is infinite. Yesterday he had enumerated the nodes of the tree in such a way:
Let's call the nodes' level a number of nodes that occur on the way to this node from the root, including this node. This way, only the root has the level
equal to 1, while only its two sons has the level equal to 2.
Then, let's take all the nodes with the odd level and enumerate them with consecutive odd numbers, starting from the smallest levels and the leftmost nodes,
going to the rightmost nodes and the highest levels.
Then, let's take all the nodes with the even level and enumerate them with consecutive even numbers, starting from the smallest levels and the leftmost nodes,
going to the rightmost nodes and the highest levels.
Here you can see the visualization of the process. For example, in odd levels, the root was enumerated first, then, there were enumerated roots' left sons' sons and roots' right sons' sons.
You are given the string of symbols, let's call it S. Each symbol is either l or r. Naturally, this sequence denotes some path from the root, where l means going
to the left son and r means going to the right son.
Please, help Chef to determine the number of the last node in this path.
Input
The first line contains single integer T number of test cases.
Each of next T lines contain a string S consisting only of the symbols l and r.
Output
Per each line output the number of the last node in the path, described by S, modulo 109+7.
Constraints
1 ≤ |T| ≤ 5
1 ≤ |S| ≤ 10^5
Remember that the tree is infinite, so each path described by appropriate S is a correct one.
Example
Input:
4
lrl
rll
r
lllr Output:
10
14
4
13
Explanation
See the example in the statement for better understanding the samples.
题解
找找数字间的规律即可。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 1000000007
using namespace std;
int T,n;
char a[100010];
ll s;
void ji(int i)
{
if(a=='l') s=(s<<1)%mod;
else s=((s<<1)+2)%mod;
}
void ou(int i)
{
if(a=='l') s=((s<<1)-1)%mod;
else s=((s<<1)+1)%mod;
}
int main()
{
scanf("%d",&T);
while(T--)
{scanf("%s",a+1);
n=strlen(a+1); s=1;
for(int i=1;i<=n;i++)
{if(i&1) ji(i);
else ou(i);
}
printf("%lld\n",s);
}
return 0;
}