#include <bits/stdc++.h>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef pair<int,int> PII;
const int N = 200010;
const int inf = 1e9;
int n,Q,b[N],sum0[N],sum1[N];
PII p[N];
struct Node{
	int l,r,mx;
}t[N*4];
void build(int x,int l,int r){
	t[x].l=l,t[x].r=r;
	t[x].mx = -inf;
	if(t[x].l==t[x].r) return;
	int mid = l+r>>1;
	build(x<<1,l,mid);
	build(x<<1|1,mid+1,r);
}
void change(int  x,int p,int v){
	t[x].mx = max(t[x].mx,v);
	if(t[x].l==t[x].r) return;
	int mid = t[x].l+t[x].r>>1;
	if(p<=mid) change(x<<1,p,v);
	else change(x<<1|1,p,v);
}
int query(int x,int l,int r){
	//cout<<x<<" "<<l<<' '<<r<<' '<<t[x].l<<" "<<t[x].r<<endl;;
	if(t[x].l==l && t[x].r==r) return t[x].mx;
	int mid = t[x].l+t[x].r>>1;
	if(r<=mid) return query(x<<1,l,r);
	else if(l>mid) return query(x<<1|1,l,r);
	else return max(query(x<<1,l,mid),query(x<<1|1,mid+1,r));
}
int main(){
	scanf("%d",&n);
	for(int i = 1;i<=n;i++)scanf("%d",b+i);
	for(int i =1;i<=n;i++){
		sum0[i] =sum0[i-1];
		sum1[i] = sum1[i-1];
		if(b[i]==0) sum0[i]++;
		else sum1[i]++;
	}
	build(1,0,n);
	scanf("%d",&Q);
	for(int i = 1;i<=Q;i++)scanf("%d%d",&p[i].X,&p[i].Y);
	sort(p+1,p+Q+1);
	change(1,0,0);
	int mx = 0,ans = 0;
	for(int i =1,j = 1;i<=n;i++){
		while(j<=Q && p[j].X==i){
			change(1,p[j].Y,mx+sum0[i-1]-sum1[i-1]);
			int tmp = query(1,i,p[j].Y);
			change(1,p[j].Y,tmp);
			j++;
		}
		ans = max(ans,query(1,i,i)+sum1[i]);
		ans = max(ans,mx+sum0[i]);
		//cout<<i<<endl;
		
		mx = max(mx,query(1,i,i)+sum1[i]-sum0[i]);
	//	cout<<ans<<endl;
	}
	printf("%d\n",n-ans);
	return 0;
} 
### Prompt

Your task is to create a Cpp solution to the following problem:

You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.

You can perform Q kinds of operations. The i-th operation is as follows:

* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.



Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.

Constraints

* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.

Input

Input is given from Standard Input in the following format:


N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q


Output

Print the minimum possible hamming distance.

Examples

Input

3
1 0 1
1
1 3


Output

1


Input

3
1 0 1
2
1 1
3 3


Output

0


Input

3
1 0 1
2
1 1
2 3


Output

1


Input

5
0 1 0 1 0
1
1 5


Output

2


Input

9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7


Output

3


Input

15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13


Output

5


Input

10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9


Output

1

### Response

```cpp
#include <bits/stdc++.h>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef pair<int,int> PII;
const int N = 200010;
const int inf = 1e9;
int n,Q,b[N],sum0[N],sum1[N];
PII p[N];
struct Node{
	int l,r,mx;
}t[N*4];
void build(int x,int l,int r){
	t[x].l=l,t[x].r=r;
	t[x].mx = -inf;
	if(t[x].l==t[x].r) return;
	int mid = l+r>>1;
	build(x<<1,l,mid);
	build(x<<1|1,mid+1,r);
}
void change(int  x,int p,int v){
	t[x].mx = max(t[x].mx,v);
	if(t[x].l==t[x].r) return;
	int mid = t[x].l+t[x].r>>1;
	if(p<=mid) change(x<<1,p,v);
	else change(x<<1|1,p,v);
}
int query(int x,int l,int r){
	//cout<<x<<" "<<l<<' '<<r<<' '<<t[x].l<<" "<<t[x].r<<endl;;
	if(t[x].l==l && t[x].r==r) return t[x].mx;
	int mid = t[x].l+t[x].r>>1;
	if(r<=mid) return query(x<<1,l,r);
	else if(l>mid) return query(x<<1|1,l,r);
	else return max(query(x<<1,l,mid),query(x<<1|1,mid+1,r));
}
int main(){
	scanf("%d",&n);
	for(int i = 1;i<=n;i++)scanf("%d",b+i);
	for(int i =1;i<=n;i++){
		sum0[i] =sum0[i-1];
		sum1[i] = sum1[i-1];
		if(b[i]==0) sum0[i]++;
		else sum1[i]++;
	}
	build(1,0,n);
	scanf("%d",&Q);
	for(int i = 1;i<=Q;i++)scanf("%d%d",&p[i].X,&p[i].Y);
	sort(p+1,p+Q+1);
	change(1,0,0);
	int mx = 0,ans = 0;
	for(int i =1,j = 1;i<=n;i++){
		while(j<=Q && p[j].X==i){
			change(1,p[j].Y,mx+sum0[i-1]-sum1[i-1]);
			int tmp = query(1,i,p[j].Y);
			change(1,p[j].Y,tmp);
			j++;
		}
		ans = max(ans,query(1,i,i)+sum1[i]);
		ans = max(ans,mx+sum0[i]);
		//cout<<i<<endl;
		
		mx = max(mx,query(1,i,i)+sum1[i]-sum0[i]);
	//	cout<<ans<<endl;
	}
	printf("%d\n",n-ans);
	return 0;
} 
```