CodeForces

You are given a string s and should processm queries. Each query is described by two 1-based indicesl_i,r_i and integerk_i. It means that you should cyclically shift the substrings[l_i...r_i] k_i times. The queries should be processed one after another in the order they are given.

One operation of a cyclic shift (rotation) is equivalent to moving the last character to the position of the first character and shifting all other characters one position to the right.

For example, if the string s is abacaba and the query is l₁ = 3, r₁ = 6, k₁ = 1 then the answer isabbacaa. If after that we would process the queryl₂ = 1, r₂ = 4, k₂ = 2 then we would get the stringbaabcaa.

Input

The first line of the input contains the string s (1 ≤ |s| ≤ 10 000) in its initial state, where|s| stands for the length of s. It contains only lowercase English letters.

Second line contains a single integer m (1 ≤ m ≤ 300) — the number of queries.

The i-th of the next m lines contains three integers l_i,r_i andk_i (1 ≤ l_i ≤ r_i ≤ |s|, 1 ≤ k_i ≤ 1 000 000) — the description of the i-th query.

Output

Print the resulting string s after processing allm queries.

Example

Input

abacaba23 6 11 4 2

Output

baabcaa

虽然说是水题，但我看题看了好久才看懂。。。

题目大意就是给你个字符串，执行若干次操作后输出该字符串

每次操作就是将给定区间[l,r]的所有字母首尾相连形成一个环，然后环上每个元素向后移动k位（其实就是循环移动，只是我不会表述）

代码

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<cmath>using namespace std;int pos[10005];//用该数组保存该位置对应字母在char[]数组中的位置int prepos[10005];//用该数组保存该位置之前对应字母在char[]数组中的位置char t[10005];int main(){    int m,l,r,k;    while(cin>>t)    {        int n=strlen(t);        for(int i = 1; i <= n; i++) {                pos[i]=i;                prepos[i]=i;        }        cin>>m;        for(int i = 0; i < m; i++)        {            cin>>l>>r>>k;            for(int j = r; j >= l; j--)            {                int temp=l+(j-l+k)%(r-l+1);                pos[temp] = prepos[j];            }            for(int j = r; j >= l; j--)                prepos[j]=pos[j];        }        for(int i =1; i <= n; i++)                cout<<t[pos[i]-1];            cout<<endl;    }    return 0;}