#include<bits/stdc++.h>
#define N 200005
#define inf 1000000000
using namespace std;
struct node {
	int x,y;
}a[N];
int i,j,k,l,n,m,ans,d[N*4][4],f[N*4],s[2][N],x;
inline bool cmp(const node x,const node y) {
	return x.y<y.y;
}
inline void down(int o,int l,int r) {
	for (int i=1;i<=2;i++) {
		f[o*2]=min(f[o*2],d[o][i]+s[i-1][l]);
		d[o*2][i]=min(d[o*2][i],d[o][i]);
		f[o*2+1]=min(f[o*2+1],d[o][i]+s[i-1][r]);
		d[o*2+1][i]=min(d[o*2+1][i],d[o][i]);
		d[o][i]=inf;
	}
}
inline void build(int o,int l,int r) {
	d[o][1]=d[o][2]=inf;
	if (l==r) {
		f[o]=s[1][l];
		return ;
	}
	int mid=(l+r)>>1;
	build(o*2,l,mid),build(o*2+1,mid+1,r);
}
inline void change(int o,int l,int r,int ll,int rr,int w,int p) {
	if (ll>rr) return ;
	if (l==ll&&r==rr) {
		f[o]=min(f[o],s[w-1][l]+p);
		d[o][w]=min(d[o][w],p);
		return ;
	}
	down(o,l,r);
	int mid=(l+r)>>1;
	if (rr<=mid) change(o*2,l,mid,ll,rr,w,p);
	else if (ll>mid) change(o*2+1,mid+1,r,ll,rr,w,p);
	else change(o*2,l,mid,ll,mid,w,p),change(o*2+1,mid+1,r,mid+1,rr,w,p);
}
inline int ask(int o,int l,int r,int ll) {
	if (ll<=0) return 0;
	if (l==ll&&r==ll) return f[o];
	down(o,l,r);
	int mid=(l+r)>>1;
	if (ll<=mid) return ask(o*2,l,mid,ll);
	else return ask(o*2+1,mid+1,r,ll);
}
int main() {
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&x),s[x][i]=s[x][i-1]+1,s[x^1][i]=s[x^1][i-1];
	scanf("%d",&m);
	for (i=1;i<=m;i++) scanf("%d%d",&a[i].x,&a[i].y);
	sort(a+1,a+1+m,cmp);
	j=1;
	build(1,1,n);
	for (i=1;i<=n;i++) {
		while (a[j].y==i) {
			x=ask(1,1,n,a[j].x-1)-s[0][a[j].x-1];
			change(1,1,n,a[j].x,a[j].y,1,x);
			x=ask(1,1,n,a[j].y)-s[1][a[j].y];
			change(1,1,n,a[j].y+1,n,2,x);
			j++;
		}
	}
	printf("%d\n",ask(1,1,n,n));
}
### Prompt

Create a solution in Cpp for 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 N 200005
#define inf 1000000000
using namespace std;
struct node {
	int x,y;
}a[N];
int i,j,k,l,n,m,ans,d[N*4][4],f[N*4],s[2][N],x;
inline bool cmp(const node x,const node y) {
	return x.y<y.y;
}
inline void down(int o,int l,int r) {
	for (int i=1;i<=2;i++) {
		f[o*2]=min(f[o*2],d[o][i]+s[i-1][l]);
		d[o*2][i]=min(d[o*2][i],d[o][i]);
		f[o*2+1]=min(f[o*2+1],d[o][i]+s[i-1][r]);
		d[o*2+1][i]=min(d[o*2+1][i],d[o][i]);
		d[o][i]=inf;
	}
}
inline void build(int o,int l,int r) {
	d[o][1]=d[o][2]=inf;
	if (l==r) {
		f[o]=s[1][l];
		return ;
	}
	int mid=(l+r)>>1;
	build(o*2,l,mid),build(o*2+1,mid+1,r);
}
inline void change(int o,int l,int r,int ll,int rr,int w,int p) {
	if (ll>rr) return ;
	if (l==ll&&r==rr) {
		f[o]=min(f[o],s[w-1][l]+p);
		d[o][w]=min(d[o][w],p);
		return ;
	}
	down(o,l,r);
	int mid=(l+r)>>1;
	if (rr<=mid) change(o*2,l,mid,ll,rr,w,p);
	else if (ll>mid) change(o*2+1,mid+1,r,ll,rr,w,p);
	else change(o*2,l,mid,ll,mid,w,p),change(o*2+1,mid+1,r,mid+1,rr,w,p);
}
inline int ask(int o,int l,int r,int ll) {
	if (ll<=0) return 0;
	if (l==ll&&r==ll) return f[o];
	down(o,l,r);
	int mid=(l+r)>>1;
	if (ll<=mid) return ask(o*2,l,mid,ll);
	else return ask(o*2+1,mid+1,r,ll);
}
int main() {
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&x),s[x][i]=s[x][i-1]+1,s[x^1][i]=s[x^1][i-1];
	scanf("%d",&m);
	for (i=1;i<=m;i++) scanf("%d%d",&a[i].x,&a[i].y);
	sort(a+1,a+1+m,cmp);
	j=1;
	build(1,1,n);
	for (i=1;i<=n;i++) {
		while (a[j].y==i) {
			x=ask(1,1,n,a[j].x-1)-s[0][a[j].x-1];
			change(1,1,n,a[j].x,a[j].y,1,x);
			x=ask(1,1,n,a[j].y)-s[1][a[j].y];
			change(1,1,n,a[j].y+1,n,2,x);
			j++;
		}
	}
	printf("%d\n",ask(1,1,n,n));
}
```