#include<cstdio>
#include<cstring>
struct vec{
	int x,nx;
}e[5000001];
int h[200001],p,cnt,co[200001];
void ae(int fr,int to){
	e[++p]=(vec){to,h[fr]};h[fr]=p; 
}
int mi[200001<<2];
int n,f[200001],q;
int Min(int p,int q){
	return p<q?p:q;
}
int Max(int p,int q){
	return p>q?p:q;
}
int ask(int i,int l,int r,int ql,int qr){
	if(r<ql||l>qr) return 2147483647;
	if(ql<=l&&r<=qr) return mi[i];
	int m=(l+r)>>1;
	return Min(ask(i<<1,l,m,ql,qr),ask(i<<1|1,m+1,r,ql,qr));
}
void upd(int i,int l,int r,int k,int pos){
	if(r<pos||l>pos) return;
	if(l==r&&r==pos){
		mi[i]=Min(mi[i],k);return;
	}
	int m=(l+r)>>1;
	upd(i<<1,l,m,k,pos);upd(i<<1|1,m+1,r,k,pos);
	mi[i]=Min(mi[i<<1],mi[i<<1|1]);
}
int main(){
	scanf("%d",&n);
	for(int i=1,a;i<=n;i++){
		scanf("%d",&a);
		cnt+=!a;
		co[i]=a?1:-1;
	}
	scanf("%d",&q);
	for(int i=1,l,r;i<=q;i++){
		scanf("%d%d",&l,&r);
		ae(l,r);
	}
	memset(mi,0x3f,sizeof(mi));
	memset(f,0x3f,sizeof(f));
	f[0]=0;
	for(int i=1;i<=n;i++){
		for(int j=h[i];j;j=e[j].nx){
			int mi=f[i-1];
			mi=Min(mi,ask(1,1,n,Max(i-1,1),e[j].x));
			if(mi<f[e[j].x]){
				f[e[j].x]=mi;
				upd(1,1,n,mi,e[j].x);
			}
		}
		f[i]=Min(f[i],f[i-1]+co[i]);
	}
	printf("%d\n",f[n]+cnt);
}
### 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<cstdio>
#include<cstring>
struct vec{
	int x,nx;
}e[5000001];
int h[200001],p,cnt,co[200001];
void ae(int fr,int to){
	e[++p]=(vec){to,h[fr]};h[fr]=p; 
}
int mi[200001<<2];
int n,f[200001],q;
int Min(int p,int q){
	return p<q?p:q;
}
int Max(int p,int q){
	return p>q?p:q;
}
int ask(int i,int l,int r,int ql,int qr){
	if(r<ql||l>qr) return 2147483647;
	if(ql<=l&&r<=qr) return mi[i];
	int m=(l+r)>>1;
	return Min(ask(i<<1,l,m,ql,qr),ask(i<<1|1,m+1,r,ql,qr));
}
void upd(int i,int l,int r,int k,int pos){
	if(r<pos||l>pos) return;
	if(l==r&&r==pos){
		mi[i]=Min(mi[i],k);return;
	}
	int m=(l+r)>>1;
	upd(i<<1,l,m,k,pos);upd(i<<1|1,m+1,r,k,pos);
	mi[i]=Min(mi[i<<1],mi[i<<1|1]);
}
int main(){
	scanf("%d",&n);
	for(int i=1,a;i<=n;i++){
		scanf("%d",&a);
		cnt+=!a;
		co[i]=a?1:-1;
	}
	scanf("%d",&q);
	for(int i=1,l,r;i<=q;i++){
		scanf("%d%d",&l,&r);
		ae(l,r);
	}
	memset(mi,0x3f,sizeof(mi));
	memset(f,0x3f,sizeof(f));
	f[0]=0;
	for(int i=1;i<=n;i++){
		for(int j=h[i];j;j=e[j].nx){
			int mi=f[i-1];
			mi=Min(mi,ask(1,1,n,Max(i-1,1),e[j].x));
			if(mi<f[e[j].x]){
				f[e[j].x]=mi;
				upd(1,1,n,mi,e[j].x);
			}
		}
		f[i]=Min(f[i],f[i-1]+co[i]);
	}
	printf("%d\n",f[n]+cnt);
}
```