Problem: Almost periodic string

Given the string S. Answer Q queries for this string S. In the i-th query, determine if S [l_i, \ r_i] can be a string of period t_i when you can change from S [l_i, \ r_i] to one character. S [l, \ r] represents the substring of the string S from the lth to the rth.

If the character string W is a character string with period t, then for i \ = \ 1, \ 2, \ ..., \ | W | − t, W_ {i} = W_ {i + t}. I will do it.

Constraints

* 1 ≤ | S | ≤ 10 ^ 5
* 1 ≤ Q ≤ 10 ^ 5
* 1 ≤ l_i ≤ r_i ≤ | S |
* 1 ≤ t_i ≤ r_i − l_i + 1
* S consists of only lowercase letters



Input Format

Input is given from standard input in the following format.


S
Q
l_1 r_1 t_1
...
l_Q r_Q t_Q


Output Format

Output over Q lines. On line i, print the answer to the i-th query with `Yes` or` No`.

Sample Input 1


abcabcaxcabc
Four
1 9 3
8 12 3
1 4 2
2 3 2


Sample Output 1


Yes
Yes
No
Yes


Sample Input 2


isuruu
Four
3 6 1
3 6 2
3 6 3
2 4 1


Sample Output 2


Yes
Yes
Yes
No






Example

Input

abcabcaxcabc
4
1 9 3
8 12 3
1 4 2
2 3 2


Output

Yes
Yes
No
Yes

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define ll long long
#define INF 1000000005
#define MOD 1000000007
#define EPS 1e-10
#define rep(i,n) for(int i=0;i<(int)(n);++i)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;--i)
#define srep(i,s,t) for(int i=(int)(s);i<(int)(t);++i)
#define each(a,b) for(auto (a): (b))
#define all(v) (v).begin(),(v).end()
#define len(v) (int)(v).size()
#define zip(v) sort(all(v)),v.erase(unique(all(v)),v.end())
#define cmx(x,y) x=max(x,y)
#define cmn(x,y) x=min(x,y)
#define fi first
#define se second
#define pb push_back
#define show(x) cout<<#x<<" = "<<(x)<<endl
#define spair(p) cout<<#p<<": "<<p.fi<<" "<<p.se<<endl
#define svec(v) cout<<#v<<":";rep(kbrni,v.size())cout<<" "<<v[kbrni];cout<<endl
#define sset(s) cout<<#s<<":";each(kbrni,s)cout<<" "<<kbrni;cout<<endl
#define smap(m) cout<<#m<<":";each(kbrni,m)cout<<" {"<<kbrni.first<<":"<<kbrni.second<<"}";cout<<endl

using namespace std;

typedef pair<int,int> P;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<double> vd;
typedef vector<P> vp;
typedef vector<string> vs;

const int MAX_N = 100005;

struct RollingHash {
	static const ll mo0=1000000007,mo1=1000000009; static ll mul0,mul1;
    static vector<ll> pmo[2];
	string s; int sz; vector<ll> hash_[2];
	void init(string s) {
		this->s=s; sz=(int)s.size();
		hash_[0].resize(sz+1,0),hash_[1].resize(sz+1,0);
		if(!mul0) mul0=10009+(((ll)&mul0)>>5)%259,mul1=10007+(((ll)&mul1)>>5)%257;
		if(pmo[0].empty()) pmo[0].pb(1),pmo[1].pb(1);
		rep(i,sz) hash_[0][i+1]=(hash_[0][i]*mul0+s[i])%mo0;	//hash_[0][i]はインデックス0~i-1までの文字列のハッシュ値
		rep(i,sz) hash_[1][i+1]=(hash_[1][i]*mul1+s[i])%mo1;
	}
	pair<ll,ll> hash(int l,int r){	//文字列sのインデックス[l,r]までの部分文字列のハッシュ値
		if(l>r) return make_pair(0,0);
		while((int)pmo[0].size()<r+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
		return make_pair((hash_[0][r+1]+(mo0-hash_[0][l]*pmo[0][r+1-l]%mo0))%mo0,
			             (hash_[1][r+1]+(mo1-hash_[1][l]*pmo[1][r+1-l]%mo1))%mo1);
	}
    pair<ll,ll> update_hash(int l, int r, int k, char c){  //文字列sのs[k]をcに変えたときの[l,r]までの部分文字列のハッシュ値
        if(l > k || r < k) return hash(l,r);
        while((int)pmo[0].size()<r+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
        return make_pair((hash_[0][r+1]+(mo0-hash_[0][l]*pmo[0][r+1-l]%mo0)+pmo[0][r-k]*(c - s[k]+mo0))%mo0,
                         (hash_[1][r+1]+(mo1-hash_[1][l]*pmo[1][r+1-l]%mo1)+pmo[1][r-k]*(c - s[k]+mo1))%mo1);
    }
	pair<ll,ll> hash(string s) { init(s); return hash(0,(int)s.size()-1); }	//文字列s全体のハッシュ値
	static pair<ll,ll> concat(pair<ll,ll> L,pair<ll,ll> R,int RL) { //文字列L+Rのハッシュ値,RLはRの文字列の長さ
		while((int)pmo[0].size()<RL+2) pmo[0].pb(pmo[0].back()*mul0%mo0), pmo[1].pb(pmo[1].back()*mul1%mo1);
		return make_pair((R.first + L.first*pmo[0][RL])%mo0,(R.second + L.second*pmo[1][RL])%mo1);
	}
};
vector<ll> RollingHash::pmo[2]; ll RollingHash::mul0,RollingHash::mul1;

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s;
    cin >> s;
    RollingHash rh;
    rh.init(s);
    int q;
    cin >> q;
    rep(i,q){
        int x,y,z;
        cin >> x >> y >> z;
        --x,--y;
        pll p1 = rh.hash(x,y-z), p2 = rh.hash(x+z,y);
        if(p1 == p2){
            cout << "Yes\n";
            continue;
        }
        int l = 0,r = y-x-z+1;
        while(r-l>1){
            int mid = (l+r)/2;
            if(rh.hash(x,x+mid-1) == rh.hash(x+z,x+z+mid-1)){
                l = mid;
            }else{
                r = mid;
            }
        }
        if(rh.update_hash(x,y-z,x+l,s[x+z+l]) == rh.update_hash(x+z,y,x+l,s[x+z+l])){
            cout << "Yes\n";
        }else{
            if(rh.update_hash(x,y-z,x+z+l,s[x+l]) == rh.update_hash(x+z,y,x+z+l,s[x+l])){
                cout << "Yes\n";
            }else{
                cout << "No\n";
            }
        }
    }
    return 0;
}