#include<bits/stdc++.h>
using namespace std;
#define REP(i,st,ed) for(int i=(int)(st),i##end=(int)(ed);i<=i##end;++i)
#define DREP(i,st,ed) for(int i=(int)(st),i##end=(int)(ed);i>=i##end;--i)
template<typename T>bool chkmin(T &x,const T &y){return x>y?x=y,1:0;}
template<typename T>bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<typename T>T read(){
	T x=0,f=1;
	char c=getchar();
	while((c<'0')||(c>'9')){if(c=='-')f=-1;c=getchar();}
	while((c>='0')&&(c<='9'))x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
#define read() read<int>()
template<typename T>void write(T x,char c){
	static char t[25];
	static int tlen;
	t[tlen=1]=c;
	if(x<0)putchar('-'),x=-x;
	do t[++tlen]=(x%10)^48;
	while(x/=10);
	while(tlen)putchar(t[tlen--]);
}
const int maxn=200005;
int n,m;
int a[maxn],dp[maxn];
vector<int>g[maxn];
struct seg_tree{
	int Min[maxn<<2];
	void update(int p,int l,int r,int L,int R,int x){
		if((L<=l)&&(R>=r)){chkmin(Min[p],x);return;}
		int mid=(l+r)>>1,u=p<<1,v=u+1;
		if(L<=mid)update(u,l,mid,L,R,x);
		if(R>mid)update(v,mid+1,r,L,R,x);
	}
	int query(int p,int l,int r,int pos){
		if(l==r)return Min[p];
		int mid=(l+r)>>1,u=p<<1,v=u+1,res=Min[p];
		if(pos<=mid)chkmin(res,query(u,l,mid,pos));
		else chkmin(res,query(v,mid+1,r,pos));
		return res;
	}
}T;
int main(){
	n=read();
	REP(i,1,n)a[i]=a[i-1]+(read()^1);
	m=read();
	REP(i,1,m){
		int l=read(),r=read();
		g[r].push_back(l-1);
	}
	memset(T.Min,0x3f,sizeof(T.Min));
	memset(dp,0x3f,sizeof(dp));
	dp[0]=0;
	REP(i,1,n){
		dp[i]=dp[i-1]+(a[i]-a[i-1]==0);
		REP(j,0,g[i].size()-1){
			int tmp=min(dp[g[i][j]]-a[g[i][j]],T.query(1,0,n,g[i][j]));
//			cout<<i<<' '<<g[i][j]<<' '<<tmp+a[i]<<' '<<endl;
			chkmin(dp[i],tmp+a[i]);
			T.update(1,0,n,g[i][j]+1,i,tmp);
		}
//		cout<<i<<' '<<dp[i]<<endl;
	}
//	REP(i,1,n)
//		write(dp[i],i==n?'\n':'\n');
	write(dp[n],'\n');
	return 0;
}
### Prompt

Please create a solution in cpp 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 REP(i,st,ed) for(int i=(int)(st),i##end=(int)(ed);i<=i##end;++i)
#define DREP(i,st,ed) for(int i=(int)(st),i##end=(int)(ed);i>=i##end;--i)
template<typename T>bool chkmin(T &x,const T &y){return x>y?x=y,1:0;}
template<typename T>bool chkmax(T &x,const T &y){return x<y?x=y,1:0;}
template<typename T>T read(){
	T x=0,f=1;
	char c=getchar();
	while((c<'0')||(c>'9')){if(c=='-')f=-1;c=getchar();}
	while((c>='0')&&(c<='9'))x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return x*f;
}
#define read() read<int>()
template<typename T>void write(T x,char c){
	static char t[25];
	static int tlen;
	t[tlen=1]=c;
	if(x<0)putchar('-'),x=-x;
	do t[++tlen]=(x%10)^48;
	while(x/=10);
	while(tlen)putchar(t[tlen--]);
}
const int maxn=200005;
int n,m;
int a[maxn],dp[maxn];
vector<int>g[maxn];
struct seg_tree{
	int Min[maxn<<2];
	void update(int p,int l,int r,int L,int R,int x){
		if((L<=l)&&(R>=r)){chkmin(Min[p],x);return;}
		int mid=(l+r)>>1,u=p<<1,v=u+1;
		if(L<=mid)update(u,l,mid,L,R,x);
		if(R>mid)update(v,mid+1,r,L,R,x);
	}
	int query(int p,int l,int r,int pos){
		if(l==r)return Min[p];
		int mid=(l+r)>>1,u=p<<1,v=u+1,res=Min[p];
		if(pos<=mid)chkmin(res,query(u,l,mid,pos));
		else chkmin(res,query(v,mid+1,r,pos));
		return res;
	}
}T;
int main(){
	n=read();
	REP(i,1,n)a[i]=a[i-1]+(read()^1);
	m=read();
	REP(i,1,m){
		int l=read(),r=read();
		g[r].push_back(l-1);
	}
	memset(T.Min,0x3f,sizeof(T.Min));
	memset(dp,0x3f,sizeof(dp));
	dp[0]=0;
	REP(i,1,n){
		dp[i]=dp[i-1]+(a[i]-a[i-1]==0);
		REP(j,0,g[i].size()-1){
			int tmp=min(dp[g[i][j]]-a[g[i][j]],T.query(1,0,n,g[i][j]));
//			cout<<i<<' '<<g[i][j]<<' '<<tmp+a[i]<<' '<<endl;
			chkmin(dp[i],tmp+a[i]);
			T.update(1,0,n,g[i][j]+1,i,tmp);
		}
//		cout<<i<<' '<<dp[i]<<endl;
	}
//	REP(i,1,n)
//		write(dp[i],i==n?'\n':'\n');
	write(dp[n],'\n');
	return 0;
}
```