#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,m,ans,f[1000010],id[500010],lz[1000010],a[500010],b[500010];
vector<int>zh[500010];
void build(int x,int l,int r){
	if(l==r){
		if(l==0)f[x]=0;
		 else f[x]=-1e8;
		return;
	}
	int mid=(l+r)/2;
	build(x*2,l,mid);build(x*2+1,mid+1,r);
	f[x]=max(f[x*2],f[x*2+1]);
}
void pushdown(int x){
	if(lz[x]){
		lz[x*2]+=lz[x];lz[x*2+1]+=lz[x];
		f[x*2]+=lz[x];f[x*2+1]+=lz[x];
		lz[x]=0;
	}
}
void ins(int x,int l,int r,int t,int k,int w){
	if(l==t&&r==k){
		f[x]+=w;lz[x]+=w;
		return;
	}
	int mid=(l+r)/2;
	pushdown(x);
	if(mid>=k)ins(x*2,l,mid,t,k,w);
	 else if(t>mid)ins(x*2+1,mid+1,r,t,k,w);
	  else ins(x*2,l,mid,t,mid,w),ins(x*2+1,mid+1,r,mid+1,k,w);
	f[x]=max(f[x*2],f[x*2+1]);
}
void xiu(int x,int l,int r,int t,int w){
	if(l==r){
		f[x]=max(f[x],w);
		return;
	}
	int mid=(l+r)/2;
	pushdown(x);
	if(mid>=t)xiu(x*2,l,mid,t,w);
	 else xiu(x*2+1,mid+1,r,t,w);
	f[x]=max(f[x*2],f[x*2+1]);
}
int query(int x,int l,int r,int t,int k){
	if(l==t&&r==k)return f[x];
	pushdown(x);
	int mid=(l+r)/2;
	if(mid>=k)return query(x*2,l,mid,t,k);
	 else if(t>mid)return query(x*2+1,mid+1,r,t,k);
	  else return max(query(x*2,l,mid,t,mid),query(x*2+1,mid+1,r,mid+1,k));
}
//void get(int x,int l,int r,int t,int k)
int main(){
	int i,j,xlh,x;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&x);
		id[i]=x;
		if(x==1)b[i]=1;
		b[i]+=b[i-1];
	}
	scanf("%d",&m);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&a[i]);
		zh[x].push_back(i);
	}
	build(1,0,n);
	for(i=1;i<=n;i++){
		//ins(1,0,n,0,i-1,(id[i]==0));
	//	printf("%d %d\n",i,query(1,0,n,0,i-1));
		for(j=0;j<zh[i].size();j++){
			x=zh[i][j];
			x=a[x];
			xlh=query(1,0,n,i,x)+b[x];
			xlh=max(xlh,query(1,0,n,0,i-1)+b[x]-b[i-1]);
			//printf("%d %d %d\n",i,x,xlh);
			xiu(1,0,n,x,xlh-b[x]);
		}
		ins(1,0,n,0,i-1,(id[i]==0));
		ins(1,0,n,i,i,b[i]);
		//ans=max(ans,f[1]);
	}
	ans=f[1];
	printf("%d",n-ans);
}
### 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<bits/stdc++.h>
#define ll long long
using namespace std;
int n,m,ans,f[1000010],id[500010],lz[1000010],a[500010],b[500010];
vector<int>zh[500010];
void build(int x,int l,int r){
	if(l==r){
		if(l==0)f[x]=0;
		 else f[x]=-1e8;
		return;
	}
	int mid=(l+r)/2;
	build(x*2,l,mid);build(x*2+1,mid+1,r);
	f[x]=max(f[x*2],f[x*2+1]);
}
void pushdown(int x){
	if(lz[x]){
		lz[x*2]+=lz[x];lz[x*2+1]+=lz[x];
		f[x*2]+=lz[x];f[x*2+1]+=lz[x];
		lz[x]=0;
	}
}
void ins(int x,int l,int r,int t,int k,int w){
	if(l==t&&r==k){
		f[x]+=w;lz[x]+=w;
		return;
	}
	int mid=(l+r)/2;
	pushdown(x);
	if(mid>=k)ins(x*2,l,mid,t,k,w);
	 else if(t>mid)ins(x*2+1,mid+1,r,t,k,w);
	  else ins(x*2,l,mid,t,mid,w),ins(x*2+1,mid+1,r,mid+1,k,w);
	f[x]=max(f[x*2],f[x*2+1]);
}
void xiu(int x,int l,int r,int t,int w){
	if(l==r){
		f[x]=max(f[x],w);
		return;
	}
	int mid=(l+r)/2;
	pushdown(x);
	if(mid>=t)xiu(x*2,l,mid,t,w);
	 else xiu(x*2+1,mid+1,r,t,w);
	f[x]=max(f[x*2],f[x*2+1]);
}
int query(int x,int l,int r,int t,int k){
	if(l==t&&r==k)return f[x];
	pushdown(x);
	int mid=(l+r)/2;
	if(mid>=k)return query(x*2,l,mid,t,k);
	 else if(t>mid)return query(x*2+1,mid+1,r,t,k);
	  else return max(query(x*2,l,mid,t,mid),query(x*2+1,mid+1,r,mid+1,k));
}
//void get(int x,int l,int r,int t,int k)
int main(){
	int i,j,xlh,x;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&x);
		id[i]=x;
		if(x==1)b[i]=1;
		b[i]+=b[i-1];
	}
	scanf("%d",&m);
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&a[i]);
		zh[x].push_back(i);
	}
	build(1,0,n);
	for(i=1;i<=n;i++){
		//ins(1,0,n,0,i-1,(id[i]==0));
	//	printf("%d %d\n",i,query(1,0,n,0,i-1));
		for(j=0;j<zh[i].size();j++){
			x=zh[i][j];
			x=a[x];
			xlh=query(1,0,n,i,x)+b[x];
			xlh=max(xlh,query(1,0,n,0,i-1)+b[x]-b[i-1]);
			//printf("%d %d %d\n",i,x,xlh);
			xiu(1,0,n,x,xlh-b[x]);
		}
		ins(1,0,n,0,i-1,(id[i]==0));
		ins(1,0,n,i,i,b[i]);
		//ans=max(ans,f[1]);
	}
	ans=f[1];
	printf("%d",n-ans);
}
```