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>
#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;

class suffixarray{
public:
    int sz,index1,index2;
    vector<int> rnk,tmp,sa,lcp;
    string recs;
    suffixarray(string s){
        recs = s;
        sz = (int)s.size();
        rnk.resize(sz+1),tmp.resize(sz+1);
        make_sa();
    }
    void make_sa(){
        index1 = sz;
        sa.resize(index1+1);
        rep(i,index1+1){
            sa[i] = i;
            rnk[i] = i<index1?recs[i]:-1;
        }
        auto comp = [&](int i,int j){
            if(rnk[i] != rnk[j]){
                return rnk[i] < rnk[j];
            }else{
                int ri = (i+index2<=index1)?rnk[i+index2]:-1;
                int rj = (j+index2<=index1)?rnk[j+index2]:-1;
                return ri < rj;
            }
        };
        for(index2=1;index2<=index1;index2*=2){
            sort(sa.begin(),sa.end(),comp);
            tmp[sa[0]] = 0;
            for(int i=1;i<=index1;i++){
                tmp[sa[i]] = tmp[sa[i-1]]+(comp(sa[i-1],sa[i])?1:0);
            }
            rep(i,index1+1){
                rnk[i] = tmp[i];
            }
        }
    }
    //lcp[i]???sa[i]??¨sa[i+1]???????????±?????\???????????\???
    //??????????????????lca??????????????????????????????????????????sa???unzip????????°??????
    void make_lcp(){
        lcp.resize(sz+1);
        rep(i,sz+1){
            rnk[sa[i]] = i;
        }
        int h = 0;
        lcp[0] = 0;
        rep(i,sz){
            int j = sa[rnk[i]-1];
            if(h > 0){
                h--;
            }
            for(;j+h<sz&&i+h<sz;h++){
                if(recs[j+h] != recs[i+h]){
                    break;
                }
            }
            lcp[rnk[i]-1] = h;
        }
    }
};

template<typename V> class segtree {
private:
    int n,sz;
    vector<V> node;
public:
    segtree(vector<V>& v){
        sz = (int)v.size();
        n = 1;
        while(n < sz){
            n *= 2;
        }
        node.resize(2*n-1);
        rep(i,sz){
            node[i+n-1] = v[i];
        }
        for(int i=n-2; i>=0; i--){
            node[i] = min(node[i*2+1],node[i*2+2]);
        }
    }
    void update(int k,V a)
    {
    	k += n-1;
    	node[k] = a;
    	while(k>0){
    		k = (k-1)/2;
    		node[k] = min(node[2*k+1],node[2*k+2]);
    	}
    }
    V query(int a,int b,int k=0,int l=0,int r=-1)
    {
        if(r < 0) r = n;
    	if(r <= a || b <= l){
    		return numeric_limits<V>::max();
    	}
    	if(a <= l && r <= b){
    		return node[k];
    	}else{
    		V vl = query(a,b,2*k+1,l,(l+r)/2);
    		V vr = query(a,b,2*k+2,(l+r)/2,r);
    		return min(vl,vr);
    	}
    }
    void print()
    {
        rep(i,sz){
            cout << query(i,i+1) << " ";
        }
        cout << endl;
    }
};

int stsearch(suffixarray& ag,string& cri)
{
    int n = ag.sz;
    int l=0,r=n+1;
    vi& sa = ag.sa;
    while(r-l>1){
        int mid = (l+r)/2;
        if(n-sa[mid] < len(cri)){
            string t = ag.recs.substr(sa[mid],n-sa[mid]);
            if(t <= cri){
                l = mid;
            }else{
                r = mid;
            }
        }else{
            string t = ag.recs.substr(sa[mid],len(cri));
            if(t < cri){
                l = mid;
            }else{
                r = mid;
            }
        }
    }
    return r;
}

int edsearch(suffixarray& ag,string& cri)
{
    int n = ag.sz;
    int l=0,r=n+1;
    vi& sa = ag.sa;
    while(r-l>1){
        int mid = (l+r)/2;
        if(n-sa[mid] < len(cri)){
            string t = ag.recs.substr(sa[mid],n-sa[mid]);
            if(t < cri){
                l = mid;
            }else{
                r = mid;
            }
        }else{
            string t = ag.recs.substr(sa[mid],len(cri));
            if(t <= cri){
                l = mid;
            }else{
                r = mid;
            }
        }
    }
    return r;
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    string s;
    cin >> s;
    suffixarray sa(s);
    vi tmp(len(sa.sa));
    rep(i,len(tmp)){
        tmp[i] = -sa.sa[i];
    }
    // rep(i,len(sa.sa)){
    //     cout << i << " " << s.substr(sa.sa[i],len(s)-sa.sa[i]) << "\n";
    // }
    segtree<int> s1(sa.sa),s2(tmp);
    int m;
    cin >> m;
    rep(i,m){
        string x,y;
        cin >> x >> y;
        int r1 = stsearch(sa,x),r2 = edsearch(sa,x);
        int r3 = stsearch(sa,y),r4 = edsearch(sa,y);
        // cout << r1 << " " << r2 << " " << r3 << " " << r4 << "\n";
        if(r1 >= r2 || r3 >= r4){
            cout << "0\n";
            continue;
        }
        int v1 = s1.query(r1,r2),v2 = -s2.query(r3,r4);
        if(v1+len(x)-len(y) > v2){
            cout << "0\n";
        }else{
            cout << v2+len(y)-v1 << "\n";
        }
    }
    return 0;
}