为什么 $O(n^2)$ 的哈希能过？

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int mod1=1e9+123,mod2=5e7+17;
const int base1=131,base2=61;
int pw1[1000005],pw2[1000005];
int hsh1[1000005],hsh2[1000005];
int border[1000005];
int n,len;
string s;
pair<int,int> hsh(int l,int r){
	int h1=hsh1[r]-hsh1[l-1]*pw1[r-l+1]%mod1;
	h1=(h1%mod1+mod1)%mod1;
	int h2=hsh2[r]-hsh2[l-1]*pw2[r-l+1]%mod2;
	h2=(h2%mod2+mod2)%mod2;
	return {h1,h2};
}
bool same(int a,int b,int c,int d){
	return hsh(a,b)==hsh(c,d);
}
int f(int x,int y){
	if(x>y) swap(x,y);
	if(s[x]!=s[y]) return 0;
	int l=1,r=x-1,mx=x-1;
	while(l<=r){
		int mid=(l+r)/2;
		if(same(x-mid+1,x,y-mid+1,y)){
			l=mid+1;
			mx=mid;
		}else{
			r=mid-1;
		}
	}
	mx=min(mx,border[x]);
	for(int i=mx;i>=1;i--){
		if(same(1,i,x-i+1,x)&&same(1,i,y-i+1,y)){
			return i;
		}
	}
	return 0;
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>s;
	len=s.length();
	s="#"+s;
	pw1[0]=pw2[0]=1;
	int j=0;
	for(int i=2;i<=len;i++){
		while(j>0&&s[i]!=s[j+1]) j=border[j];
		if(s[i]==s[j+1]) j++;
		border[i]=j;
	}
	for(int i=1;i<=len;i++){
		pw1[i]=pw1[i-1]*base1%mod1;
		pw2[i]=pw2[i-1]*base2%mod2;
	}
	for(int i=1;i<=len;i++){
		hsh1[i]=(hsh1[i-1]*base1+s[i])%mod1;
		hsh2[i]=(hsh2[i-1]*base2+s[i])%mod2;
	}
	cin>>n;
	while(n--){
		int a,b;
		cin>>a>>b;
		cout<<f(a,b)<<"\n";
	}
	return 0;
}