#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=200100;
int n,a[maxn],q;
struct node
{
	int l,r;
}c[maxn];
bool cmp(node a,node b)
{
	if(a.l==b.l) return a.r>b.r;
	return a.l<b.l; 
}
struct tree
{
	int l,r,mi;
	tree *ls,*rs;
	tree()
	{
		ls=rs=NULL;
		l=r=mi=0;
	}
	void update()
	{
		mi=min(ls->mi,rs->mi);
	}
	void build(int lx,int rx)
	{
		l=lx;r=rx;
		if(l==r) {mi=(l==0)?0:maxn;return;}
		int mid=(l+r)>>1;
		(ls=new tree)->build(lx,mid);
		(rs=new tree)->build(mid+1,rx);
		update();
	}
	void mdf(int pl,int c)
	{
		if(l==r) {mi=min(mi,c);return;}
		int mid=(l+r)>>1;
		if(pl<=mid) ls->mdf(pl,c);
		else rs->mdf(pl,c);
		update();
	}
	int qry(int lx,int rx)
	{
		if(l==lx&&r==rx) return mi;
		int mid=(l+r)>>1;
		if(rx<=mid) return ls->qry(lx,rx);
		else if(lx>mid) return rs->qry(lx,rx);
		else return min(ls->qry(lx,mid),rs->qry(mid+1,rx));
	}
}*xtr;
int main()
{
	scanf("%d",&n);
	int num=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);	
		num+=(1-a[i]);
	}
	(xtr=new tree)->build(0,n);	
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
		scanf("%d%d",&c[i].l,&c[i].r);	
	sort(c+1,c+q+1,cmp);	
	c[0].l=1;
	for(int i=1;i<=q;i++)
	{
		for(int j=c[i-1].l;j<c[i].l;j++)
			xtr->mdf(j,xtr->qry(j-1,j-1)+2*a[j]-1);
		xtr->mdf(c[i].r,xtr->qry(c[i].l-1,c[i].r));	
	}
	for(int j=c[q].l;j<=n;j++)
		xtr->mdf(j,xtr->qry(j-1,j-1)+2*a[j]-1);
	printf("%d",num+xtr->qry(n,n));
	return 0;
}

### Prompt

Please formulate 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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=200100;
int n,a[maxn],q;
struct node
{
	int l,r;
}c[maxn];
bool cmp(node a,node b)
{
	if(a.l==b.l) return a.r>b.r;
	return a.l<b.l; 
}
struct tree
{
	int l,r,mi;
	tree *ls,*rs;
	tree()
	{
		ls=rs=NULL;
		l=r=mi=0;
	}
	void update()
	{
		mi=min(ls->mi,rs->mi);
	}
	void build(int lx,int rx)
	{
		l=lx;r=rx;
		if(l==r) {mi=(l==0)?0:maxn;return;}
		int mid=(l+r)>>1;
		(ls=new tree)->build(lx,mid);
		(rs=new tree)->build(mid+1,rx);
		update();
	}
	void mdf(int pl,int c)
	{
		if(l==r) {mi=min(mi,c);return;}
		int mid=(l+r)>>1;
		if(pl<=mid) ls->mdf(pl,c);
		else rs->mdf(pl,c);
		update();
	}
	int qry(int lx,int rx)
	{
		if(l==lx&&r==rx) return mi;
		int mid=(l+r)>>1;
		if(rx<=mid) return ls->qry(lx,rx);
		else if(lx>mid) return rs->qry(lx,rx);
		else return min(ls->qry(lx,mid),rs->qry(mid+1,rx));
	}
}*xtr;
int main()
{
	scanf("%d",&n);
	int num=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);	
		num+=(1-a[i]);
	}
	(xtr=new tree)->build(0,n);	
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
		scanf("%d%d",&c[i].l,&c[i].r);	
	sort(c+1,c+q+1,cmp);	
	c[0].l=1;
	for(int i=1;i<=q;i++)
	{
		for(int j=c[i-1].l;j<c[i].l;j++)
			xtr->mdf(j,xtr->qry(j-1,j-1)+2*a[j]-1);
		xtr->mdf(c[i].r,xtr->qry(c[i].l-1,c[i].r));	
	}
	for(int j=c[q].l;j<=n;j++)
		xtr->mdf(j,xtr->qry(j-1,j-1)+2*a[j]-1);
	printf("%d",num+xtr->qry(n,n));
	return 0;
}

```