Problem: Given the string S and m queries. The i-th query is given by the two strings xi and yi.

For each query, answer the longest substring of the string S, starting with xi and ending with yi.

For the string S, | S | represents the length of S. Also, the fact that the character string T is a substring of the character string S means that a certain integer i exists and Tj = Si + j is satisfied for 1 ≤ j ≤ | T |. Where Tj represents the jth character of T.

Constraints

* 1 ≤ | S | ≤ 2 x 105
* 1 ≤ m ≤ 105
* 1 ≤ | xi |, | yi |
* $ \ sum ^ m_ {i = 1} $ (| xi | + | yi |) ≤ 2 x 105
* S and xi, yi consist only of half-width lowercase letters.

Input

The input is given in the following format.


S
m
x1 y1
x2 y2
::
xm ym


* The character string S is given on the first line.
* The number of queries m is given in the second line.
* Of the m lines from the 3rd line, the i-th query string xi, yi is given on the i-th line, separated by spaces.

Output

Answer the maximum substring length in the following format.


len1
len2
::
lenm


Output the longest substring length leni that satisfies the condition for the i-th query on the i-th line of the m lines from the first line. If there is no such substring, output 0.

Examples

Input

abracadabra
5
ab a
a a
b c
ac ca
z z


Output

11
11
4
3
0


Input

howistheprogress
4
ist prog
s ss
how is
the progress


Output

9
12
5
11


Input

icpcsummertraining
9
mm m
icpc summer
train ing
summer mm
i c
i i
g g
train i
summer er


Output

2
10
8
0
4
16
1
6
6

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
using Int = long long;
#define MAX_N 400050
int N,dat[2][2*MAX_N-1];
void init(int n_){
  N=1;
  while(N<n_) N*=2;
  for(int i=0;i<2*N-1;i++) {
    dat[0][i]=INT_MAX;
    dat[1][i]=INT_MIN;
  }
}
void update(int k,int a){
  k+=N-1;
  dat[0][k]=dat[1][k]=a;
  while(k>0){
    k=(k-1)/2;
    dat[0][k]=min(dat[0][k*2+1],dat[0][k*2+2]);
    dat[1][k]=max(dat[1][k*2+1],dat[1][k*2+2]);
  }
}
int query(int a,int b,int k,int l,int r,int f){
  if(r<=a||b<=l) return f?INT_MIN:INT_MAX;
  if(a<=l&&r<=b) return dat[f][k];
  else{
    int vl=query(a,b,k*2+1,l,(l+r)/2,f);
    int vr=query(a,b,k*2+2,(l+r)/2,r,f);
    return f?max(vl,vr):min(vl,vr);
  }
}


struct SuffixArray{
  int n,k;
  string S;
  vector<int> sa,lcp;
  SuffixArray(){}
  SuffixArray(string& S):S(S){init();}
  void init(){
    n=S.length();
    S.push_back('$');
    build_sa();
    build_lcp();
    build_rmq();
  }
  void build_sa(){
    sa.clear();
    sa.resize(n+1,0);
    iota(sa.begin(),sa.end(),0);
    sort(sa.begin(),sa.end(),
	 [&](int a,int b){
	   if(S[a]==S[b]) return a>b;
	   return S[a]<S[b];
	 });
    vector<int> c(n+1,0),r(n+1),cnt(n+1),s(n+1);
    for(int i=0;i<=n;i++) r[i]=S[i];
    for(int len=1;len<=n;len*=2){
      for(int i=0;i<=n;i++){
	 c[sa[i]]=
	   i>0 &&
	   r[sa[i-1]]==r[sa[i]] &&
	   sa[i-1]+len<=n &&
	   r[sa[i-1]+len/2]==r[sa[i]+len/2] ?
	   c[sa[i-1]]:i;
      }
      iota(cnt.begin(),cnt.end(),0);
      copy(sa.begin(),sa.end(),r.begin());
      for(int i=0;i<=n;i++){
	int s1=r[i]-len;
	if(s1>=0) sa[cnt[c[s1]]++]=s1;
      }
      c.swap(r);
    }
  }
  
  bool lt_substr(string &T,int si=0,int ti=0){
    int sn=S.size(),tn=T.size();
    while(si<sn&&ti<tn){
      if(S[si]<T[ti]) return 1;
      if(S[si]>T[ti]) return 0;
      si++;ti++;
    }
    return si>=sn&&ti<tn;
  }
  
  int lower_bound(string& T){
    int low=0,high=n+1;
    while(low+1<high){
      int mid=(low+high)/2;
      if(lt_substr(T,sa[mid],0)) low=mid;
      else high=mid;
    }
    return high;
  }
  
  int upper_bound(string& T){
    T.back()++;
    int res=lower_bound(T);
    T.back()--;
    return res;
  }
  
  // O(|T|*log|S|)
  int count(string& T){
    return upper_bound(T)-lower_bound(T);
  }
  
  void build_lcp(){
    lcp.clear();
    lcp.resize(n+1,0);
    vector<int> r2(n+1);
    for(int i=0;i<=n;i++) r2[sa[i]]=i;
    int h=0;
    lcp[0]=0;
    for(int i=0;i<n;i++){
      int j=sa[r2[i]-1];
      if(h>0) h--;
      for(;j+h<n&&i+h<n;h++){
	if(S[j+h]!=S[i+h]) break;
      }
      lcp[r2[i]-1]=h;
    }
  }
  
  int getlcp(int p,string &T,int d){
    int i=0;
    int len=min((int)T.length()-d,(int)S.length()-p-d);
    while(i<len&&S[p+d+i]==T[d+i]) i++;
    return i;
  }

  struct RMQ{
    int n;
    vector<int> dat;
    const int def=INT_MAX;
    RMQ(){}
    void init(int n_){
      n=1;
      while(n<n_) n*=2;
      dat.clear();
      dat.resize(2*n-1,def);
    }
    void build(int n_, vector<int>& a){
      for(int i=0;i<n_;i++) dat[i+n-1]=a[i];
      for(int i=n-2;i>=0;i--)
	dat[i]=min(dat[i*2+1],dat[i*2+2]);
    }
    int query(int a,int b,int k,int l,int r){
      if(r<=a||b<=l) return def;
      if(a<=l&&r<=b) return dat[k];
      else{
	int vl=query(a,b,k*2+1,l,(l+r)/2);
	int vr=query(a,b,k*2+2,(l+r)/2,r);
	return min(vl,vr);
      }
    }
    int query(int a,int b){
      return query(max(a,0),b,0,0,n);
    }
  };
  
  RMQ rmq;
  void build_rmq(){
    rmq.init(n);
    rmq.build(n,lcp);
  }

  int lower_bound2(string &T){
    int sl=S.length(),tl=T.length();
    if(lt_substr(T,sa[sl-1],0)) return sl;
    int p=tl;
    int low=0,high=sl-1;
    int l=getlcp(sa[low],T,0);
    int r=getlcp(sa[high],T,0);
    while(low+1<high){
      int mid=(low+high)/2;
      int k;
      if(l>=r){
	int m=rmq.query(low,mid);
	if(m<l){
	  high=mid,r=m;
	  continue;
	}
	k=l+getlcp(sa[mid],T,l);
      }else{
	int m=rmq.query(mid,high);
	if(m<r){
	  low=mid,l=m;
	  continue;
	}
	k=r+getlcp(sa[mid],T,r);
      }
      if(k==p) high=mid,r=k;
      else if(S[sa[mid]+k]<T[k]) low=mid,l=k;
      else high=mid,r=k;
    }
				 
    return high;
  }

  int upper_bound2(string &T){
    T.back()++;
    int res=lower_bound2(T);
    T.back()--;
    return res;
  }
  
  // O(|T|+log|S|)
  int count2(string& T){
    return  upper_bound2(T)-lower_bound2(T);
  }
};

int main(){
  string S;
  int m;
  cin>>S>>m;
  init(S.size()+2);
  SuffixArray sa(S);
  for(int i=0;i<=(int)S.size();i++){
    //cout<<i<<endl;
    update(i,sa.sa[i]);
    //cout<<i<<":"<<sa.sa[i]<<endl;
  }
  //cout<<"OK"<<endl;
  for(int i=0;i<m;i++){
    string x,y;cin>>x>>y;
    int lx=sa.lower_bound2(x);
    int rx=sa.upper_bound2(x);
    int ly=sa.lower_bound2(y);
    int ry=sa.upper_bound2(y);
    int ans;
    //cout<<lx<<" "<<rx<<endl<<ly<<" "<<ry<<endl<<query(ly,ry,0,0,N,1)<<":"<<query(lx,rx,0,0,N,0)<<endl;
    if(rx-lx<=0||ry-ly<=0) ans=0;
    else{
      int a=query(ly,ry,0,0,N,1),b=query(lx,rx,0,0,N,0);
      if(b+(int)x.size()>a+(int)y.size()) ans=0;
      else ans=a-b+y.size();
    }
    cout<<ans<<endl;
  }
  return 0;
}