#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
struct segtree{
	#define s (1<<19)
	int seg[s];
	void init(){
		fill(seg,seg+s,-INF);
	}
	void update(int k,int a){
		k+=s/2-1; seg[k]=max(seg[k],a);
		while(k>0){
			k=(k-1)/2;
			seg[k]=max(seg[k*2+1],seg[k*2+2]);
		}
	}
	int query(int a,int b,int k,int l,int r){
		if(r<a || b<l || b<a) return -INF;
		if(a<=l && r<=b) return seg[k];
		else{
			int vl=query(a,b,k*2+1,l,(l+r)/2);
			int vr=query(a,b,k*2+2,(l+r)/2+1,r);
			return max(vl,vr);
		}
	}
}seg1,seg2;
int n,q,b[200005],f[200005];
P p[200005];
int main(){
	scanf("%d",&n);
	repn(i,n){
		scanf("%d",&b[i]);
		f[i] = f[i-1]+(b[i]==1?1:-1);
	}
	scanf("%d",&q);
	repn(i,q){
		int L,R; scanf("%d%d",&L,&R);
		p[i] = mp(L,R);
	}
	sort(p+1,p+q+1);
	seg1.init(); seg2.init();
	int ret = 0;
	seg1.update(0,0); seg2.update(0,0);
	repn(i,q){
		int L = p[i].fi,R = p[i].sc;
		//[1,...,L-1]
		int x = seg1.query(0,L-1,0,0,(1<<18)-1);
		int dp = -INF;
		x += f[R]-f[L-1];
		dp = max(dp,x);
		x = seg2.query(L,R-1,0,0,(1<<18)-1);
		x += f[R];
		dp = max(dp,x);
		assert(dp>=-1e8);
	//	cout<<dp<<endl;
		seg1.update(R,dp);
	//	cout<<R<<dp<<endl;
		seg2.update(R,dp-f[R]);
	}
	ret = max(ret,seg1.query(0,(1<<18)-1,0,0,(1<<18)-1));
	ret *= -1;
	//cout<<ret<<endl;
	for(int i=1;i<=n;i++){
		ret += b[i];
	}
	cout<<ret<<endl;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(int i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
struct segtree{
	#define s (1<<19)
	int seg[s];
	void init(){
		fill(seg,seg+s,-INF);
	}
	void update(int k,int a){
		k+=s/2-1; seg[k]=max(seg[k],a);
		while(k>0){
			k=(k-1)/2;
			seg[k]=max(seg[k*2+1],seg[k*2+2]);
		}
	}
	int query(int a,int b,int k,int l,int r){
		if(r<a || b<l || b<a) return -INF;
		if(a<=l && r<=b) return seg[k];
		else{
			int vl=query(a,b,k*2+1,l,(l+r)/2);
			int vr=query(a,b,k*2+2,(l+r)/2+1,r);
			return max(vl,vr);
		}
	}
}seg1,seg2;
int n,q,b[200005],f[200005];
P p[200005];
int main(){
	scanf("%d",&n);
	repn(i,n){
		scanf("%d",&b[i]);
		f[i] = f[i-1]+(b[i]==1?1:-1);
	}
	scanf("%d",&q);
	repn(i,q){
		int L,R; scanf("%d%d",&L,&R);
		p[i] = mp(L,R);
	}
	sort(p+1,p+q+1);
	seg1.init(); seg2.init();
	int ret = 0;
	seg1.update(0,0); seg2.update(0,0);
	repn(i,q){
		int L = p[i].fi,R = p[i].sc;
		//[1,...,L-1]
		int x = seg1.query(0,L-1,0,0,(1<<18)-1);
		int dp = -INF;
		x += f[R]-f[L-1];
		dp = max(dp,x);
		x = seg2.query(L,R-1,0,0,(1<<18)-1);
		x += f[R];
		dp = max(dp,x);
		assert(dp>=-1e8);
	//	cout<<dp<<endl;
		seg1.update(R,dp);
	//	cout<<R<<dp<<endl;
		seg2.update(R,dp-f[R]);
	}
	ret = max(ret,seg1.query(0,(1<<18)-1,0,0,(1<<18)-1));
	ret *= -1;
	//cout<<ret<<endl;
	for(int i=1;i<=n;i++){
		ret += b[i];
	}
	cout<<ret<<endl;
}
```