#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=1<<28;
struct trnode{
	int lc,rc,c;
}tr[400010];int tot=0;
struct node{
	int l,r;
}c[200010];
int n,m,a[200010];
int bt(int l,int r)
{
	int x=++tot;
	if(l==r) tr[x].c=(l==0)?0:inf;
	if(l!=r)
	{
		int mid=(l+r)/2;
		tr[x].lc=bt(l,mid);
		tr[x].rc=bt(mid+1,r);
		tr[x].c=min(tr[tr[x].lc].c,tr[tr[x].rc].c);
	}
	return x;
}
void change(int x,int l,int r,int k,int c)
{
	if(l==r) {tr[x].c=min(tr[x].c,c);return;}
	int mid=(l+r)/2;
	if(k<=mid) change(tr[x].lc,l,mid,k,c);
	else change(tr[x].rc,mid+1,r,k,c);
	tr[x].c=min(tr[tr[x].lc].c,tr[tr[x].rc].c);
}
int findans(int x,int l,int r,int fl,int fr)
{
	if(fl==l&&r==fr) return tr[x].c;
	int mid=(l+r)/2;
	if(fr<=mid) return findans(tr[x].lc,l,mid,fl,fr);
	if(fl>mid) return findans(tr[x].rc,mid+1,r,fl,fr);
	return min(findans(tr[x].lc,l,mid,fl,mid),findans(tr[x].rc,mid+1,r,mid+1,fr));
}
bool cmp(node a,node b) {return a.l==b.l?a.r>b.r:a.l<b.l;}
int main()
{
	scanf("%d",&n);
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		ans+=1-a[i];
	}
	bt(0,n);
	scanf("%d",&m);
	for(int i=1;i<=m;i++) scanf("%d %d",&c[i].l,&c[i].r);
	sort(c+1,c+m+1,cmp);
	int p=1;
	for(int i=1;i<=n;i++)
	{
		while(c[p].l==i) change(1,0,n,c[p].r,findans(1,0,n,c[p].l-1,c[p].r)),p++;
		change(1,0,n,i,findans(1,0,n,i-1,i-1)+2*a[i]-1);
	}
	printf("%d",ans+findans(1,0,n,n,n));
}
### 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<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int inf=1<<28;
struct trnode{
	int lc,rc,c;
}tr[400010];int tot=0;
struct node{
	int l,r;
}c[200010];
int n,m,a[200010];
int bt(int l,int r)
{
	int x=++tot;
	if(l==r) tr[x].c=(l==0)?0:inf;
	if(l!=r)
	{
		int mid=(l+r)/2;
		tr[x].lc=bt(l,mid);
		tr[x].rc=bt(mid+1,r);
		tr[x].c=min(tr[tr[x].lc].c,tr[tr[x].rc].c);
	}
	return x;
}
void change(int x,int l,int r,int k,int c)
{
	if(l==r) {tr[x].c=min(tr[x].c,c);return;}
	int mid=(l+r)/2;
	if(k<=mid) change(tr[x].lc,l,mid,k,c);
	else change(tr[x].rc,mid+1,r,k,c);
	tr[x].c=min(tr[tr[x].lc].c,tr[tr[x].rc].c);
}
int findans(int x,int l,int r,int fl,int fr)
{
	if(fl==l&&r==fr) return tr[x].c;
	int mid=(l+r)/2;
	if(fr<=mid) return findans(tr[x].lc,l,mid,fl,fr);
	if(fl>mid) return findans(tr[x].rc,mid+1,r,fl,fr);
	return min(findans(tr[x].lc,l,mid,fl,mid),findans(tr[x].rc,mid+1,r,mid+1,fr));
}
bool cmp(node a,node b) {return a.l==b.l?a.r>b.r:a.l<b.l;}
int main()
{
	scanf("%d",&n);
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		ans+=1-a[i];
	}
	bt(0,n);
	scanf("%d",&m);
	for(int i=1;i<=m;i++) scanf("%d %d",&c[i].l,&c[i].r);
	sort(c+1,c+m+1,cmp);
	int p=1;
	for(int i=1;i<=n;i++)
	{
		while(c[p].l==i) change(1,0,n,c[p].r,findans(1,0,n,c[p].l-1,c[p].r)),p++;
		change(1,0,n,i,findans(1,0,n,i-1,i-1)+2*a[i]-1);
	}
	printf("%d",ans+findans(1,0,n,n,n));
}
```