#include<bits/stdc++.h>
using namespace std;
const int N=200100;
int n,m,a[N],mn[N<<2];
struct st{int l,r;}q[N];
bool cmp(st a,st b){return a.l==b.l?a.r>b.r:a.l<b.l;}
void build(int x,int l,int r)
{
	if(l==r){mn[x]=(l==0)?0:N;return;}
	int mid=l+r>>1;build(x<<1,l,mid);build(x<<1|1,mid+1,r);mn[x]=min(mn[x<<1],mn[x<<1|1]);
}
void chg(int x,int l,int r,int p,int v)
{
	if(l==r){mn[x]=min(mn[x],v);return;}
	int mid=l+r>>1;
	if(p<=mid)chg(x<<1,l,mid,p,v);else chg(x<<1|1,mid+1,r,p,v);
	mn[x]=min(mn[x<<1],mn[x<<1|1]);
}
int qry(int x,int l,int r,int tl,int tr)
{
	if(tl<=l&&r<=tr)return mn[x];
	int mid=l+r>>1;
	if(tr<=mid)return qry(x<<1,l,mid,tl,tr);
	else if(tl>mid)return qry(x<<1|1,mid+1,r,tl,tr);
	else return min(qry(x<<1,l,mid,tl,tr),qry(x<<1|1,mid+1,r,tl,tr));
}
int main()
{
	scanf("%d",&n);int s=0;
	for(int i=1;i<=n;i++){scanf("%d",&a[i]);s+=(1-a[i]);}
	build(1,0,n);scanf("%d",&m);
	for(int i=1;i<=m;i++)scanf("%d%d",&q[i].l,&q[i].r);
	sort(q+1,q+m+1,cmp);
	for(int i=1,j=1;i<=n;i++)
	{
		for(;q[j].l==i;j++)chg(1,0,n,q[j].r,qry(1,0,n,q[j].l-1,q[j].r));
		chg(1,0,n,i,qry(1,0,n,i-1,i-1)+2*a[i]-1);
	}
	printf("%d\n",s+qry(1,0,n,n,n));
	return 0;
}

### Prompt

Create a solution in cpp for 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>
using namespace std;
const int N=200100;
int n,m,a[N],mn[N<<2];
struct st{int l,r;}q[N];
bool cmp(st a,st b){return a.l==b.l?a.r>b.r:a.l<b.l;}
void build(int x,int l,int r)
{
	if(l==r){mn[x]=(l==0)?0:N;return;}
	int mid=l+r>>1;build(x<<1,l,mid);build(x<<1|1,mid+1,r);mn[x]=min(mn[x<<1],mn[x<<1|1]);
}
void chg(int x,int l,int r,int p,int v)
{
	if(l==r){mn[x]=min(mn[x],v);return;}
	int mid=l+r>>1;
	if(p<=mid)chg(x<<1,l,mid,p,v);else chg(x<<1|1,mid+1,r,p,v);
	mn[x]=min(mn[x<<1],mn[x<<1|1]);
}
int qry(int x,int l,int r,int tl,int tr)
{
	if(tl<=l&&r<=tr)return mn[x];
	int mid=l+r>>1;
	if(tr<=mid)return qry(x<<1,l,mid,tl,tr);
	else if(tl>mid)return qry(x<<1|1,mid+1,r,tl,tr);
	else return min(qry(x<<1,l,mid,tl,tr),qry(x<<1|1,mid+1,r,tl,tr));
}
int main()
{
	scanf("%d",&n);int s=0;
	for(int i=1;i<=n;i++){scanf("%d",&a[i]);s+=(1-a[i]);}
	build(1,0,n);scanf("%d",&m);
	for(int i=1;i<=m;i++)scanf("%d%d",&q[i].l,&q[i].r);
	sort(q+1,q+m+1,cmp);
	for(int i=1,j=1;i<=n;i++)
	{
		for(;q[j].l==i;j++)chg(1,0,n,q[j].r,qry(1,0,n,q[j].l-1,q[j].r));
		chg(1,0,n,i,qry(1,0,n,i-1,i-1)+2*a[i]-1);
	}
	printf("%d\n",s+qry(1,0,n,n,n));
	return 0;
}

```