#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=1000000000;
struct data{
	int l,r;
}a[410001];
struct tnode{
	int l,r,ls,rs,v;
}t[810001];
int i,j,k,n,m,b[410001],tot,ans;
void mt(int l,int r){
	int p=++tot,m;
	t[p].l=l;t[p].r=r;t[p].v=inf;
	if (l<r){
		m=(l+r)>>1;
		t[p].ls=tot+1;mt(l,m);
		t[p].rs=tot+1;mt(m+1,r);
	}else t[p].ls=t[p].rs=0;
}
void change(int p,int x,int v){
	int m=(t[p].l+t[p].r)>>1;
	t[p].v=min(v,t[p].v);
	if (!t[p].ls) return;
	change((x<=m)?t[p].ls:t[p].rs,x,v);
}
int query(int p,int l,int r){
	int res=inf,m=(t[p].l+t[p].r)>>1;
	if (l<=t[p].l&&t[p].r<=r) return t[p].v;
	if (l<=m) res=query(t[p].ls,l,r);
	if (r>=m+1) res=min(res,query(t[p].rs,l,r));
	return res;
}
bool cmp(data x,data y){
	return x.l<y.l||x.l==y.l&&x.r<y.r;
}
int main(){
//	freopen("a.in","r",stdin);
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&b[i]),ans+=1-b[i];
	tot=0;mt(1,n);scanf("%d",&m);
	for (i=1;i<=m;i++) scanf("%d%d",&a[i].l,&a[i].r);
	sort(a+1,a+1+m,cmp);
	for (i=j=1;i<=n;i++){
		for (;j<=m&&a[j].l==i;j++) change(1,a[j].r,(i>1)?query(1,i-1,a[j].r):0);//,printf("** %d\n",j);
		change(1,i,((i>1)?query(1,i-1,i-1):0)+((b[i])?1:-1));
	}
	ans=ans+query(1,n,n);
	printf("%d\n",ans);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=1000000000;
struct data{
	int l,r;
}a[410001];
struct tnode{
	int l,r,ls,rs,v;
}t[810001];
int i,j,k,n,m,b[410001],tot,ans;
void mt(int l,int r){
	int p=++tot,m;
	t[p].l=l;t[p].r=r;t[p].v=inf;
	if (l<r){
		m=(l+r)>>1;
		t[p].ls=tot+1;mt(l,m);
		t[p].rs=tot+1;mt(m+1,r);
	}else t[p].ls=t[p].rs=0;
}
void change(int p,int x,int v){
	int m=(t[p].l+t[p].r)>>1;
	t[p].v=min(v,t[p].v);
	if (!t[p].ls) return;
	change((x<=m)?t[p].ls:t[p].rs,x,v);
}
int query(int p,int l,int r){
	int res=inf,m=(t[p].l+t[p].r)>>1;
	if (l<=t[p].l&&t[p].r<=r) return t[p].v;
	if (l<=m) res=query(t[p].ls,l,r);
	if (r>=m+1) res=min(res,query(t[p].rs,l,r));
	return res;
}
bool cmp(data x,data y){
	return x.l<y.l||x.l==y.l&&x.r<y.r;
}
int main(){
//	freopen("a.in","r",stdin);
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&b[i]),ans+=1-b[i];
	tot=0;mt(1,n);scanf("%d",&m);
	for (i=1;i<=m;i++) scanf("%d%d",&a[i].l,&a[i].r);
	sort(a+1,a+1+m,cmp);
	for (i=j=1;i<=n;i++){
		for (;j<=m&&a[j].l==i;j++) change(1,a[j].r,(i>1)?query(1,i-1,a[j].r):0);//,printf("** %d\n",j);
		change(1,i,((i>1)?query(1,i-1,i-1):0)+((b[i])?1:-1));
	}
	ans=ans+query(1,n,n);
	printf("%d\n",ans);
	return 0;
}
```