#include<bits/stdc++.h>
using namespace std;
inline int read(){
	int f=1,x=0;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')
			f=-f;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=x*10+ch-'0';
		ch=getchar();
	}
	return f*x;
}
inline void work(int k){
	if(k<0){
		putchar('-');
		k=-k;
	}
	if(k>9)
		work(k/10);
	putchar(k%10+'0');
}
/*
考虑在已知字符串后面加入一个字符,如果这个字符是一个新字符,那么新添加的本质不同的字符串就是前面的长度+1
前面已经知道了,所以本质不同的字符串可以算,只需要问一次就行了
那么我们二分满足增添的个数是长度+1的位置就能够确定这个字符串是什么就行了
每次暴力SA,时间复杂度O(sum(ilog^2i)),询问次数O(logi)
*/
int n,a[1005],now=1,sa[1005],rk[1005],y[2005],tot[1005],height[1005],m;
void SA(int a[],int n){
	m=1e3;
	for(int i=1;i<=m;i++){
		rk[i]=sa[i]=tot[i]=0;
	}
	for(int i=1;i<=2*n;i++){
		y[i]=0;
	}
	for(int i=1;i<=n;i++){
		rk[i]=a[i];
		tot[rk[i]]++;
	}
	for(int i=1;i<=m;i++){
		tot[i]+=tot[i-1];
	}
	for(int i=n;i;i--){
		sa[tot[rk[i]]--]=i;
	}
	for(int k=1;k<=n;k<<=1){
		int num=0;
		for(int i=n-k+1;i<=n;i++){
			y[++num]=i;
		}
		for(int i=1;i<=n;i++){
			if(sa[i]>k)
				y[++num]=sa[i]-k;
		}
		for(int i=1;i<=m;i++){
			tot[i]=0;
		}
		for(int i=1;i<=n;i++){
			tot[rk[y[i]]]++;
		}
		for(int i=1;i<=m;i++){
			tot[i]+=tot[i-1];
		}
		for(int i=n;i;i--){
			sa[tot[rk[y[i]]]--]=y[i];
		}
		for(int i=1;i<=n;i++)[
			swap(rk[i],y[i]);
		]
		num=rk[sa[1]]=1;
		for(int i=2;i<=n;i++){
			if(y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k])
				rk[sa[i]]=num;
			else
				rk[sa[i]]=++num;
		}
		if(num==n)
			break;
		m=num;
	}
}
void LCP(int a[],int n){
	int k=0;
	for(int i=1;i<=n;i++){
		if(rk[i]==1)
			continue;
		if(k)
			k--;
		int last=rk[sa[i]-1];
		while(last+k<=n&&i+k<=n&&a[i+k]==a[last+k]){
			k++;
		}
		height[rk[i]]=k;
	}
}
int solve(int l,int r){
	SA(a+l-1,r-l+1);
	LCP(a+l-1,r-l+1);
	int sum=(r-l+1)*(r-l+2)/2;
	for(int i=1;i<=r-l+1;i++){
		sum-=height[i];
	}
	return sum;
}
int main(){
	a[1]=1;
	scanf("%d",&n);
	for(int i=2;i<=n;i++){
		int l=1,r=i;
		while(l<r){
			int mid=(l+r)>>1;
			printf("? %d %d\n",mid,i);
			fflush(stdout);
			int val=0;
			scanf("%d",&val);
			if(solve(mid,i-1)+i-mid+1==val)
				r=mid;
			else
				l=mid+1;
		}
		if(l==1)
			a[i]=++now;
		else
			a[i]=a[l-1];
	}
	return 0;
}

answer.code: In function ‘void SA(int*, int)’:
answer.code:73:40: error: expression list treated as compound expression in initializer [-fpermissive]
   73 |                         swap(rk[i],y[i]);
      |                                        ^
answer.code:73:41: error: expected ‘,’ before ‘;’ token
   73 |                         swap(rk[i],y[i]);
      |                                         ^
      |                                         ,
answer.code:73:41: error: expected identifier before ‘;’ token
answer.code:73:41: error: expected ‘]’ before ‘;’ token
   73 |                         swap(rk[i],y[i]);
      |                                         ^
      |                                         ]
answer.code: In lambda function:
answer.code:73:41: error: expected ‘{’ before ‘;’ token
answer.code: In function ‘void SA(int*, int)’:
answer.code:74:17: error: expected primary-expression before ‘]’ token
   74 |                 ]
      |                 ^
answer.code: In function ‘int main()’:
answer.code:112:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  112 |         scanf("%d",&n);
      |         ~~~~~^~~~~~~~~
answer.code:120:30: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  120 |                         scanf("%d",&val);
      |                         ~~~~~^~~~~~~~~~~