#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=2e5+10;
const int INF=1e9;
typedef pair<int,int> pii;
#define x first
#define y second
#define mp make_pair

struct node
{
	int l,r;
	node *ls,*rs;
	int mini;
} *root;
int n,a[maxn],q,pos;
int s[maxn];
pii p[maxn];
int dp[maxn];

void build(node *p,int l,int r)
{
	p->l=l;p->r=r;
	p->mini=INF;
	
	if (l==r) return;
	int mid=(l+r)>>1;
	p->ls=new node;
	p->rs=new node;
	build(p->ls,l,mid);
	build(p->rs,mid+1,r);
	return;
}

int getmin(node *p,int l,int r)
{
	if (l<=p->l && p->r<=r) return p->mini;
	else if (r<p->l || p->r<l) return INF;
	else return min(getmin(p->ls,l,r),getmin(p->rs,l,r));
}

void update(node *p,int pos,int val)
{
	if (p->l==p->r) 
	{
		p->mini=min(p->mini,val);
		return;
	}
	else
	{
		if (pos<=p->ls->r) update(p->ls,pos,val);
		else update(p->rs,pos,val);
		p->mini=min(p->ls->mini,p->rs->mini);
		return;
	}
}

int main() 
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) 
	{
		scanf("%d",&a[i]);
		s[i]=s[i-1]+(1-a[i]);
	}
	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);
	
	root=new node;
	build(root,0,n);
	for(int i=1;i<=n;i++) dp[i]=INF;
	update(root,0,0);
	dp[0]=0;
	
	pos=1;
	for(int i=1;i<=n;i++)
	{
		dp[i]=min(dp[i],dp[i-1]+a[i]);
		
		int st=pos;
		while(pos<=q && p[pos].x==i) pos++;
		int ed=pos-1;
		
		for(int j=st;j<=ed;j++)
		{
			int tmp=min(dp[i-1]+s[p[j].y]-s[i-1],
				getmin(root,p[j].x,p[j].y-1)+s[p[j].y]);
			dp[p[j].y]=min(dp[p[j].y],tmp);
			update(root,p[j].y,tmp-s[p[j].y]);
		}
	}
	
	printf("%d\n",dp[n]);
	return 0;
}
### Prompt

Generate 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 <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=2e5+10;
const int INF=1e9;
typedef pair<int,int> pii;
#define x first
#define y second
#define mp make_pair

struct node
{
	int l,r;
	node *ls,*rs;
	int mini;
} *root;
int n,a[maxn],q,pos;
int s[maxn];
pii p[maxn];
int dp[maxn];

void build(node *p,int l,int r)
{
	p->l=l;p->r=r;
	p->mini=INF;
	
	if (l==r) return;
	int mid=(l+r)>>1;
	p->ls=new node;
	p->rs=new node;
	build(p->ls,l,mid);
	build(p->rs,mid+1,r);
	return;
}

int getmin(node *p,int l,int r)
{
	if (l<=p->l && p->r<=r) return p->mini;
	else if (r<p->l || p->r<l) return INF;
	else return min(getmin(p->ls,l,r),getmin(p->rs,l,r));
}

void update(node *p,int pos,int val)
{
	if (p->l==p->r) 
	{
		p->mini=min(p->mini,val);
		return;
	}
	else
	{
		if (pos<=p->ls->r) update(p->ls,pos,val);
		else update(p->rs,pos,val);
		p->mini=min(p->ls->mini,p->rs->mini);
		return;
	}
}

int main() 
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) 
	{
		scanf("%d",&a[i]);
		s[i]=s[i-1]+(1-a[i]);
	}
	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);
	
	root=new node;
	build(root,0,n);
	for(int i=1;i<=n;i++) dp[i]=INF;
	update(root,0,0);
	dp[0]=0;
	
	pos=1;
	for(int i=1;i<=n;i++)
	{
		dp[i]=min(dp[i],dp[i-1]+a[i]);
		
		int st=pos;
		while(pos<=q && p[pos].x==i) pos++;
		int ed=pos-1;
		
		for(int j=st;j<=ed;j++)
		{
			int tmp=min(dp[i-1]+s[p[j].y]-s[i-1],
				getmin(root,p[j].x,p[j].y-1)+s[p[j].y]);
			dp[p[j].y]=min(dp[p[j].y],tmp);
			update(root,p[j].y,tmp-s[p[j].y]);
		}
	}
	
	printf("%d\n",dp[n]);
	return 0;
}
```