#include<bits/stdc++.h>

using namespace std;

#define N 800000

int n,m,ans,a[N],f[N],cnt[2][N],seg[2][N];
struct node{
	int l,r;
	bool operator < (const node p) const{return l<p.l;}
}b[N];

#define ls (x<<1)
#define rs (x<<1|1)

void build(int x,int l,int r){
	if (l==r){seg[0][x]=seg[1][x]=n; return;}
	int mid=(l+r)>>1;
	build(ls,l,mid); build(rs,mid+1,r);
	seg[0][x]=min(seg[0][ls],seg[0][rs]);
	seg[1][x]=min(seg[1][ls],seg[1][rs]);
}

void mdf(int x,int l,int r,int k,int t){
	if (l==r){
		seg[0][x]=min(seg[0][x],t-cnt[0][k]);
		seg[1][x]=min(seg[1][x],t-cnt[1][k]);
		return;
	}
	int mid=(l+r)>>1;
	if (k<=mid) mdf(ls,l,mid,k,t); else mdf(rs,mid+1,r,k,t);
	seg[0][x]=min(seg[0][ls],seg[0][rs]);
	seg[1][x]=min(seg[1][ls],seg[1][rs]);
}

int qry(int s[],int x,int l,int r,int L,int R){
	if (L>R) return n;
	if (l>=L&&r<=R) return s[x];
	int mid=(l+r)>>1,ret=n;
	if (L<=mid) ret=min(ret,qry(s,ls,l,mid,L,R));
	if (R>mid) ret=min(ret,qry(s,rs,mid+1,r,L,R));
	return ret;
}

#undef ls
#undef rs

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		cnt[0][i]=cnt[0][i-1]+(a[i]==0);
		cnt[1][i]=cnt[1][i-1]+(a[i]==1);
	}
	scanf("%d",&m);
	for (int i=1;i<=m;i++) scanf("%d%d",&b[i].l,&b[i].r);
	sort(b+1,b+m+1);
	memset(f,0x3f,sizeof f); f[0]=0;
	build(1,1,n); ans=cnt[1][n];
	for (int i=1;i<=m;i++){
		int l=b[i].l,r=b[i].r;
		f[r]=cnt[1][l-1]+cnt[0][r]-cnt[0][l-1];
		f[r]=min(f[r],qry(seg[1],1,1,n,1,l-1)+cnt[1][l-1]+cnt[0][r]-cnt[0][l-1]);
		f[r]=min(f[r],qry(seg[0],1,1,n,l,r)+cnt[0][r]);
		mdf(1,1,n,r,f[r]);
		ans=min(ans,f[r]+cnt[1][n]-cnt[1][r]);
	}
	printf("%d\n",ans);
	
	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<bits/stdc++.h>

using namespace std;

#define N 800000

int n,m,ans,a[N],f[N],cnt[2][N],seg[2][N];
struct node{
	int l,r;
	bool operator < (const node p) const{return l<p.l;}
}b[N];

#define ls (x<<1)
#define rs (x<<1|1)

void build(int x,int l,int r){
	if (l==r){seg[0][x]=seg[1][x]=n; return;}
	int mid=(l+r)>>1;
	build(ls,l,mid); build(rs,mid+1,r);
	seg[0][x]=min(seg[0][ls],seg[0][rs]);
	seg[1][x]=min(seg[1][ls],seg[1][rs]);
}

void mdf(int x,int l,int r,int k,int t){
	if (l==r){
		seg[0][x]=min(seg[0][x],t-cnt[0][k]);
		seg[1][x]=min(seg[1][x],t-cnt[1][k]);
		return;
	}
	int mid=(l+r)>>1;
	if (k<=mid) mdf(ls,l,mid,k,t); else mdf(rs,mid+1,r,k,t);
	seg[0][x]=min(seg[0][ls],seg[0][rs]);
	seg[1][x]=min(seg[1][ls],seg[1][rs]);
}

int qry(int s[],int x,int l,int r,int L,int R){
	if (L>R) return n;
	if (l>=L&&r<=R) return s[x];
	int mid=(l+r)>>1,ret=n;
	if (L<=mid) ret=min(ret,qry(s,ls,l,mid,L,R));
	if (R>mid) ret=min(ret,qry(s,rs,mid+1,r,L,R));
	return ret;
}

#undef ls
#undef rs

int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		cnt[0][i]=cnt[0][i-1]+(a[i]==0);
		cnt[1][i]=cnt[1][i-1]+(a[i]==1);
	}
	scanf("%d",&m);
	for (int i=1;i<=m;i++) scanf("%d%d",&b[i].l,&b[i].r);
	sort(b+1,b+m+1);
	memset(f,0x3f,sizeof f); f[0]=0;
	build(1,1,n); ans=cnt[1][n];
	for (int i=1;i<=m;i++){
		int l=b[i].l,r=b[i].r;
		f[r]=cnt[1][l-1]+cnt[0][r]-cnt[0][l-1];
		f[r]=min(f[r],qry(seg[1],1,1,n,1,l-1)+cnt[1][l-1]+cnt[0][r]-cnt[0][l-1]);
		f[r]=min(f[r],qry(seg[0],1,1,n,l,r)+cnt[0][r]);
		mdf(1,1,n,r,f[r]);
		ans=min(ans,f[r]+cnt[1][n]-cnt[1][r]);
	}
	printf("%d\n",ans);
	
	return 0;
}
			
```