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

Solution (in CPP):

Solution:
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define reg register
using namespace std;
const int N=2e5+5;
struct node
{
	int l,r;
	inline friend bool operator < (node a,node b)
	{
		return a.l==b.l?a.r>b.r:a.l<b.l;
	}
}q[N];
int n,m,a[N],cnt,tmin[N<<2];
inline int read()
{
	int x=0,w=1;
	char c=getchar();
	while (!isdigit(c)&&c!='-') c=getchar();
	if (c=='-') c=getchar(),w=-1;
	while (isdigit(c))
	{
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*w;
}
inline void pushup(int now)
{
	tmin[now]=min(tmin[now<<1],tmin[now<<1|1]);
}
void build(int l,int r,int now)
{
	if (l==r) {tmin[now]=(l?N:0); return;}
	int mid=(l+r)>>1;
	build(l,mid,now<<1); build(mid+1,r,now<<1|1);
	pushup(now);
}
void change(int l,int r,int now,int x,int c)
{
	if (l==r) {tmin[now]=min(tmin[now],c); return;}
	int mid=(l+r)>>1;
	if (x<=mid) change(l,mid,now<<1,x,c);
	else change(mid+1,r,now<<1|1,x,c);
	pushup(now);
}
int query(int L,int R,int l,int r,int now)
{
	if (l>R||r<L) return N;
	if (l>=L&&r<=R) return tmin[now];
	int mid=(l+r)>>1;
	if (mid>=R) return query(L,R,l,mid,now<<1);
	if (mid<L) return query(L,R,mid+1,r,now<<1|1);
	return min(query(L,mid,l,mid,now<<1),query(mid+1,R,mid+1,r,now<<1|1));
}
int main()
{
	n=read();
	for (reg int i=1;i<=n;i++) cnt+=(1-(a[i]=read()));
	m=read();
	for (reg int i=1;i<=m;i++) q[i].l=read(),q[i].r=read();
	sort(q+1,q+m+1); build(0,n,1);
	for (reg int i=1,now=1;i<=n;i++)
	{
		while (q[now].l==i) change(0,n,1,q[now].r,query(i-1,q[now].r,0,n,1)),++now;
		change(0,n,1,i,query(i-1,i-1,0,n,1)+(a[i]<<1)-1);
	}
	printf("%d\n",cnt+query(n,n,0,n,1));
	return 0;
}