#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define int long long
const int inf=0x3f3f3f3f3f3f3f3f;
int n,m,pre[200100][2],tree[1001000],ans,a[200100],f[200100],x,y;
vector<int>v[200100];
void modify(int t,int l,int r,int pos,int val){
	if(l==r){
		tree[t]=min(tree[t],val);
		return;
	}
	int mid=(l+r)/2;
	if(pos<=mid)modify(t*2,l,mid,pos,val);
	else modify(t*2+1,mid+1,r,pos,val);
	tree[t]=min(tree[t*2],tree[t*2+1]);
}
int query(int t,int l,int r,int L,int R){
	if(l>R||L>r)return inf;
	if(L<=l&&r<=R)return tree[t];
	int mid=(l+r)/2;
	return min(query(t*2,l,mid,L,R),query(t*2+1,mid+1,r,L,R));
}
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		pre[i][0]=pre[i-1][0]+(a[i]==0);
		pre[i][1]=pre[i-1][1]+(a[i]==1);
	}
	memset(f,0x3f,sizeof(f));
	scanf("%lld",&m);
	for(int i=1;i<=m;i++){
		scanf("%lld%lld",&x,&y);
		v[x].push_back(y);
	}
	f[0]=0;
	memset(tree,0x3f,sizeof(tree));
	for(int i=1;i<=n;i++){
		for(int j=0;j<v[i].size();j++){
			int now=v[i][j];
			int minn=min(f[i-1],query(1,1,n,max(i-1,1ll),now));
			if(minn<f[now]){
				f[now]=minn;
				modify(1,1,n,now,minn);
			}
		}
		f[i]=min(f[i],f[i-1]+((a[i]==1)?1:-1));
	}
	printf("%lld\n",f[n]+pre[n][0]);
}
### Prompt

Your challenge is to write 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<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define int long long
const int inf=0x3f3f3f3f3f3f3f3f;
int n,m,pre[200100][2],tree[1001000],ans,a[200100],f[200100],x,y;
vector<int>v[200100];
void modify(int t,int l,int r,int pos,int val){
	if(l==r){
		tree[t]=min(tree[t],val);
		return;
	}
	int mid=(l+r)/2;
	if(pos<=mid)modify(t*2,l,mid,pos,val);
	else modify(t*2+1,mid+1,r,pos,val);
	tree[t]=min(tree[t*2],tree[t*2+1]);
}
int query(int t,int l,int r,int L,int R){
	if(l>R||L>r)return inf;
	if(L<=l&&r<=R)return tree[t];
	int mid=(l+r)/2;
	return min(query(t*2,l,mid,L,R),query(t*2+1,mid+1,r,L,R));
}
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		pre[i][0]=pre[i-1][0]+(a[i]==0);
		pre[i][1]=pre[i-1][1]+(a[i]==1);
	}
	memset(f,0x3f,sizeof(f));
	scanf("%lld",&m);
	for(int i=1;i<=m;i++){
		scanf("%lld%lld",&x,&y);
		v[x].push_back(y);
	}
	f[0]=0;
	memset(tree,0x3f,sizeof(tree));
	for(int i=1;i<=n;i++){
		for(int j=0;j<v[i].size();j++){
			int now=v[i][j];
			int minn=min(f[i-1],query(1,1,n,max(i-1,1ll),now));
			if(minn<f[now]){
				f[now]=minn;
				modify(1,1,n,now,minn);
			}
		}
		f[i]=min(f[i],f[i-1]+((a[i]==1)?1:-1));
	}
	printf("%lld\n",f[n]+pre[n][0]);
}
```